Math and Reasoning Mix 2

Math and Reasoning Mix Questions

11.X is the only daughter of Y’s grandfather’s only son. Y’s grandfather has only one child. How is X related to Y?

A) Mother
B) Sister
C) Grandmother
D) Paternal Aunt

Ans: B) Sister
Tree diagram given below:-

12.Seeta said, “Ram’s only brother is the father of my son’s father”. How is Ram’s brother related to Seeta’s son?

A) Maternal uncle
B) Grandfather
C) Father
D) Paternal Uncle

Ans: C) Father
Tree Diagram given below :-

13.The perimeter of an isosceles (ସମଦ୍ୱିବାହୁ🔺) triangle is 50cm and its base is 24cm its area is

A) 60cm²
B) 50cm²
C) 180cm²
D) 90cm²

Ans: A) 60cm²

Perimeter= x + x + 24 = 50cm (Given)
x = 13
ଉଚ୍ଚତା ² = କର୍ଣ² – ଭୂମି ² ।
ଏଠାରେ କର୍ଣ = 13cm, ଭୂମି = 12cm ।
ଉଚ୍ଚତା = 169 – 144 = ✓25 = 5cm
Area of a ∆ = ½ × ଭୂମି × ଉଚ୍ଚତା => ½ × 24cm × 5cm = 60cm²

14.Select the related word/letters from the given alternatives.
9:8::16:?

A) 17
B) 14
C) 27
D) 23

Ans: C) 27

The logic here is 9:8
(3)²: (2)³
Similarly, 16:27
(4)²: (3)³
Hence, 27 is the correct answer.

15.Select the combination of letters that when sequentially placed in the gaps of the given letter series will complete the series.
ac_d_b_cbdd_a_bddb

A) b d a b c
B) b d b c a
C) b d c a b
D) c b d b c

Ans: A) b d a b c

ac b d d b a cbdd b a c bddb

16. A sold a pen to B at a loss of 60%. B sold the pen to C at a profit of 30%. If C pays Rs 260 for pen, then for how much (in Rs.) A bought the pen?

(A) 500
(B) 490
(C) 480
(D) 400

Ans: (A) 500

A Sold to B and loss 60%
Suppose A bought a pen of Rs.100 and sold to B Rs.40
A—–60% Loss——–B
(CP) 100——–40 (SP)
B—–30% Profit——C
(CP) 40——– 52(SP)
According to Question CP of C Rs.52 = 260
1 = 260/52
100 = 5 × 100 = 500

17. On selling an article for Rs. 462, a shopkeeper losses 40%. At what price he should sell the article in order to earn a profit of 20%?

(A) 542
(B) 356
(C) 684
(D) 924

Ans: (D) 924

60% = 462 => 1% = 462 ÷ 60 × 120 = 924

18. If FLOWER is coded as 14 and DISTANCE is coded as 18, then how will BUREAUCRAT be coded as?

(A) 22
(B) 18
(C) 20
(D) 28

Ans: (A) 22

FLOWER = 6 letters = 14
Logic Applied:-
6 × 2 + 2 = 14 (Given)

19. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?

(A) 4
(B) 10
(C) 15
(D) 16

Ans: (D) 16

L.C.M of 2,4,6,8,10,12 = 120Sec. ( Bell ୨ ମିନିଟ ପରେ ଏକା ସାଙ୍ଗରେ ବାଜିବ)
30 minutes ରେ ୧୫ ଥର ବାଜିବ ୧ ଥର ବାଜିବ = ୧୬ ଥର ବାଜିବ। (ଉତ୍ତର)

20. The greatest number of four digits which is divisible by 15, 25, 40, 75 is:

(A) 9000
(B) 9400
(C) 9600
(D) 9800

Ans: (C) 9600

L.C.M of 15,25,40,75 = 600
The gratest no. of four digits = 9999 .Now divide 9999 ÷ 600 = 399(ଭାଗଶେଷ)
=> 9999 – 399 = 9600(ଉତ୍ତର)

21. The smallest four digits number which is complete divisible by 4, 6, 8 is:

(A) 1016
(B) 1008
(C) 1024
(D) 1000

Ans: (B) 1008

L.C.M of 4,6,8 = 24
The smallest no. of four digits = 1000 .Now divide 1000 ÷ 24 = 16(ଭାଗଶେଷ)
=> 24 – 16 = 8
1000 + 8 = 1008(ଉତ୍ତର)

22. Find the least possible five-digit number when divided by 10, 12, 16 and 18 gives a reminder 27

(A) 10017
(B) 10107
(C) 10002
(D) 10002

Ans: (B) 10107

L.C.M of 12,10,16,18 = 720
The smallest no. of five digits = 10000 .Now divide 10000 ÷ 720 = 640 (ଭାଗଶେଷ)
=> 720 – 640 = 80
10000 + 80 = 10080(Completely Divisible by 10,12,16,18)
According to question remender should be 27.
=> 10080 + 27 = 10107

23. In a certain code language, ‘134’ means ‘good and tasty’, ‘478’ means ‘see good pictures’ and ‘729’ means ‘picture are faint’. Which of the following digits stands for ‘see’?

(A) 9
(B) 1
(C) 2
(D) 8

Ans: (D) 8
Find Common wards with common number.
4 – Good
7 – Picture

24. 543 : 12 :: 764 : ?

(A) 25
(B) 17
(C) 23
(D) 16

Ans: (B) 17

5 + 4 + 3 = 12
7 + 6 + 4 = 17

25. In a language FIFTY is written as CACTY, CAR as POL, TAR as TOL, How can TARIFF be written in that language?

(A) TOEFEL
(B) TOEFDD
(C) TOLADD
(D) TOLACC

Ans: (D) TOLACC

26. Based on the analogy given, find the missing number from the given option.
9:24::?: 6

(A) 3
(B) 2
(C) 1
(D) 5

Ans: (A) 3

9 × 3 – 3 = 24
3 × 3 – 3 = 6

27. There are five houses P, Q, R, S and T. P is Right of Q and T is left of R and Right of P. Q is right of S. Which house is in the middle?

(A) P
(B) Q
(C) T
(D) S

Ans: (A) P

S Q “P” T R

28. Select the missing number from the given series.
12, 36, 80, 150, ?

(A) 252
(B) 236
(C) 243
(D) 256

Ans: (A) 252

2³ + 2², 3³ + 3², 4³ + 4²,5³ + 5²,6³+6²
(OR)
2² × 3 = 12
3² × 4 = 36
4² × 5 = 80
5² × 6 = 150
6² × 7 = 252

29. 13, 25, 51, 101, 203, ?

(A) 405
(B) 407
(C) 456
(D) 025

Ans: (A) 405

25 = 13 × 2 – 1
51 = 25 × 2 + 1
101 = 51 × 2 – 1
203 = 101 × 2 + 1
405 = 203 × 2 – 1

30. 5,11,23,47,95,?

(A) 119
(B) 198
(C) 191
(D) 185

Ans: (C) 191

5 × 2 + 1 = 11
11 × 2 + 1 = 23
23 × 2 + 1 = 47
47 × 2 + 1 = 95
95 × 2 + 1 = 191

31. Rajan is sixth from the left end and Vinay is tenth from the right end in a row of boys. If there are eight boys between Rajan and Vinay, how many boys are there in the row?

(A) 19
(B) 16
(C) 8
(D) 24

Ans: (D) 24

—->6th 8 10th<-----

32. ୫^୧୭ + ୫^୧୮ + ୫^୧୯ + ୫^୨୦ କେଉଁ ସଂଖ୍ୟା ଦ୍ୱାରା ଭାଗ ହେବ ?

A) ୭
B) ୯
C) ୧୫
D) ୧୩

Ans:D) ୧୩
୫^୧୭ (୫^୦ + ୫^୧ + ୫^୨ + ୫^୩)
୫^୧୭(୧ + ୫ + ୨୫ + ୧୨୫)
୫^୧୭(୧୫୬)
୫^୧୭ × ୧୨ × ୧୩
ଏହି ସଂଖ୍ୟାଟି ୧୩ ଦ୍ୱାରା ଵିଭାଜ୍ୟ ହେବ। (ଉତ୍ତର)

33. 1 × 2 × 3 × 4 × 5 × ……× 29 × 30 ର ଗୁଣଫଳର ଶେଷ ଭାଗରେ କେତୋଟି ୦ (ଶୂନ) ଅଛି ?

A) ୭
B) ୫
C) ୧୦
D) ୯

Ans:A) ୭
୩୦ ÷ ୫ = ୬
୬ ÷ ୫ = ୧
= ୫^୭
୩୦ ÷ ୨ = ୧୫
୧୫ ÷ ୨ = ୭
୭ ÷ ୨ = ୩
୩ ÷ ୨ = ୧
= ୨^୨୬
୨^୨୬ × ୫^୭ [ ୫ × ୨ = କେବଳ ଗୋଟିଏ ଶୂନ ପାଇପାରିବା।]

34. If the sum of the first twelve odd numbers is added to the sum of the first twelve even numbers, then their sum will be_____?

A) 306
B) 604
C) 302
D) 300

Ans:D) 300
Sum of the first twelve odd numbers formula = n² =>12²=144
Sum of first twelve even numbers formula = n (n + 1) => 12 × (12 + 1) => 12 × 13 = 156
144 + 156 = 300

35. ‘A’ starts a business with an investment of Rs. 2,500. Seven months later, ‘B’ joins with ‘A’ as a partner. After a years, the profits are divided in the ratio of 3: 4. How much did ‘B’ contribute ?

A) 7000
B) 8000
C) 10000
D) 12500

Ans:B) 8000
A—>2500 × 12 = 30000
B—> x × 5 = 5x
^{2500 × 12}/_{x × 5} = ³/₄ = 8000

36. A boat man can row 36 km in downstream in 4 hours. If the speed of the current is 3 km/h, then find in what time will he be able to cover 6 km upstream?

A) 3hrs
B) 6hrs
C) 5hrs
D) 2hrs

Ans:D) 2hrs
Speed of current = 3km/hr
Downstream Speed(x + y) = 1hr = 9km/hr
Upstream Speed (x – y) => 6 – 3 = 3km/hr
Time = 6 ÷ 3 = 2hrs

37. From a point on a circular track of 12 km long, A, B and C started running in the same direction at the same time with speed of 4 km/h, 3 km/h and 2 km/h respectively. Then on the starting point, all three will meet again after:

A) 3hrs
B) 6hrs
C) 12hrs
D) 4hrs

Ans:C) 12hrs
^12km/HCF of 4,3,2 km/hrs
¹²/₁ = 12hrs

38. How much water be added to 16 litres of milk worth Rs. 6.75 per litre so that the value of the mixture will be Rs. 4.50 per litre ?

A) 8 litres
B) 18 litres
C) 12 litres
D) 7 litres

Ans:A) 8 litres
Alternate Method Cost of water will be zero.
Let “x” litres of water should be added.
Cost of milk = 16 x 6.75 = 108
According to the question,
16 × 6.75 = (16 + x) × 4.50
⇒ 108 = 4.50(16 + x)
⇒ 24 = 16 + x
⇒ x = 8
.°. 8 litres of water should be added.

39. Ratio of present age of Amit and Rakesh is 7:5. Amit will be 36 years old after 8 years. How old Rakesh is now?

A) 28 Years
B) 15 Years
C) 20 Years
D) 21 Years

Ans:C) 20 Years
Present age Ratio:
Amit : Rakesh = 7 : 5
Present age of Amit = 36 – 8 = 28
7 = 28 ବର୍ଷ⇒ 1 = 4 ବର୍ଷ ହେବ ।
.°. 5 = 5 × 4 = 20 ବର୍ଷ ହେବ ।

40. A sum of money at simple interest becomes five times in 24 years. The rate of interest per annum is :

A) 20 + ⁴⁰/₆%
B) 18 + ²/₃%
C) 16 + ²/₃%
D) None of these

Ans:C) 16 + ²/₃%
P = x (Let the principal be x.)
T = 24Years
R = ?
A = 5x
I =A – P = 5x – x = 4x
I = ^PTR/₁₀₀
4x = ^{x × 24 × r}/₁₀₀
^{= 6r}/₁₀₀⇒ r = ¹⁰⁰/₆

41. At what percent per annum a sum of Rs. 500 will amount to Rs. 864 in 3 years, if the interest is compounded annually ?

A) 20%
B) 18%
C) 16%
D) 25%

Ans:A) 20%
P = 500
T = 3 Years
R = ?
A = 864
I = 364
A = p(1 + ^r/₁₀₀)^t
864 = 500(1 + ^r/₁₀₀)³
⁸⁶⁴/₅₀₀ = (1 + ^r/₁₀₀)³
²¹⁶/₁₂₅ = (1 + ^r/₁₀₀)³
³√²¹⁶/₁₂₅= (1 + ^r/₁₀₀)
⁶/₅ = (1 + ^r/₁₀₀)
⁶/₅= ^{(100 + r}/₁₀₀)
⁶/₁ = ^{(100 + r}/₂₀)
100 + r = 120 ⇒ r = 120 – 100 = 20%

Ans: (A) Formula: (a+b)² = a² + b² + 2ab

43. The sum of two numbers is 192. If one- fourth of the one exceeds one-seventh of the other by 4, find the bigger number.

A) 112

B) 122

C) 132

D) 432

Ans:A) 112

Solution- The sum of two numbers is 192.
Let a + b = 192.
If one- fourth of the one exceeds one-seventh of the other by 4.
¼a – ⅐b = 4
⇒7a – 4b = 112
We got two equation
4 × (a + b = 192)   —Eq(1)
7a – 4b = 112    —-Eq(2)
Eq(1) & Eq(2) କୁ ଯୋଗକଲେ ପାଇବା 11a = 880 ⇒ a = 80
ଆପଣ a ର ଭେଲୁକୁ Eq(1) a + b = 192 ରେ ପ୍ରୟୋଗ କରନ୍ତୁ।

44. The sum of two numbers is 384 and their HCF is 48. What is the difference between the two number ?

A) 48

B) 96

C) 144

D) 192

Ans:B) 96

Solution- The sum of two numbers is 384.
HCF is 48 (Given).
Let the numbers be 48a , 48b

48a + 48b = 384 (According to question)
⇒48(a + b) = 384
⇒ a + b = 384 ÷ 48 = 8
(i) 1 + 7 = 8 ⇒ Numbers are 48 , 336
(ii) 2 + 6 = 8⇒ Numbers are 96 , 288
(iii) 3 + 5 = 8⇒ Numbers are 144 , 240
.°. Difference= 240 – 144 = 96

45. The proportion of two numbers is 3:5 and their L.C.M. is 225.find out the smaller number ?

A) 48

B) 45

C) 60

D) 92

Ans:B) 45

Solution-
Let the numbers be 3x & 5x
⇒ Their H.C.F is x.
L.C.M is 225(Given)
L.C.M × H.C.F = ସଂଖ୍ୟା ଦ୍ୱୟର ଗୁଣଫଳ
225x = 15x²
⇒x = 15
3x = 3 × 15 = 45

46. The ratio of two numbers is 5:4. If 40% of the first number is 12.What will be 50% of the second number?

A) 12

B) 14

C) 15

D) 18

Ans:A) 12

Solution-
Let the numbers be 5 & 4
40% of first number is 12(Given) ⇒⅖ × 5 = 12
⇒ 2 = 12
⇒1 = 6
⇒ First Number = 5 × 6 = 30 & Second Number = 4 × 6 = 24
.°. 50% of Second number will be 24 ÷ 2 = 12 (Answer)

47. Simple interest on a sum of money for 5 years is ²/₅ times the principal, the rate for simple interest is

A) 8%

B) 4%

C) 5%

D) 7.5%

Ans:A) 8%

Solution-
Let the sum of money (Principal) be p.
Time given 5 years
⇒SI = ^{p × t × r}/₁₀₀
⇒²/₅ × p = ^{p × 5 × r}/₁₀₀
.°. 5r = 40 ⇒r = 8% (Answer)
OR
²/₅ Here 2 is simple interest and 5 is Principal.
Just put the SI formula ⇒SI = ^{p × t × r}/₁₀₀

48. find the average of all factor of 60.

A) 12

B) 15

C) 14

D) 10

Ans:C) 14

Solution-:
Factors of 60 = 2² × 3¹ × 5¹
(2+1) × (1+1) × (1+1)
Factors of 60 = 3×2×2= 12(factors)
Factors are = 1,2,3,4,5,6,10,12,15,20,30 & 60.
Average = ^{1+2+3+4+5+6+10+12+15+20+30+60}/₁₂= 14

49. Average age of three boys is 22 years. If the ratio of their ages is 6: 9:7, then the age of the youngest boy is

A) 18 years

B) 15 years

C) 27 years

D) 21 years

Ans:A) 18

Solution-: Average age of 3 boys is 22 (given).
Total age of 3 boys is 22 × 3 = 66
Their ages be 6x,9x & 7x.
22x = 66 (According to question)
x = 3
.°. 6 × 3 = 18 years Answer

50.ଯଦି 60 ହେଉଛି, 10, 12, 15, x ଏବଂ y ର ହାରାହାରି ର 400% ଅଟେ, ତେବେ x ଏବଂ y ର ହାରାହାରି ଖୋଜନ୍ତୁ ।

A) 19

B) 15

C) 27

D) 20

Ans:A) 19
Solution-:^{10 + 12 + 15 + x + y}/₅ = Average × 400% = 60 (Given)⇒Average = 15
^{37 + x + y}/₅= 15
37 + x + y = 75
⇒ x + y = 38 ⇒ହାରାହାରି = 38 ÷ 2 = 19

51. 45ଟି ସଂଖ୍ୟାର ହାରାହାରି ହେଉଛି 150 । ପରେ ଏହା ଦେଖାଗଲା ଯେ ଗୋଟିଏ ସଂଖ୍ୟା 46କୁ ଭୁଲ କ୍ରମେ 91 ଭାବରେ ଲେଖା ହୋଇଛି, ତା’ପରେ ସଠିକ୍ ହାରାହାରି ନିର୍ଣ୍ଣୟ କରନ୍ତୁ ?

A) 147

B) 151

C) 149

D) 153

Ans:C) 149
Solution-: 45ଟି ସଂଖ୍ୟାର ସମଷ୍ଟି = 45 × 150 = 6750
ଗୋଟିଏ ସଂଖ୍ୟା 46କୁ ଭୁଲ କ୍ରମେ 91 ଭାବରେ ଲେଖା ହୋଇଛି ⇒ 91 – 46 = 45
⇒ 6750 – 45 = 6705
ସଠିକ୍ ହାରାହାରି = 6705 ÷ 45 = 149

52.Three friends had dinner at a restaurant. When the bill was received, Anamika paid ²/₃ as much as Vinita paid and Vinita paid ¹/₂ as much as Lalita paid. What fraction of the bill did Vinita pay?

A) ²/₁₃

B) ) ³/₁₁

C) ) ¹¹/₃

D) ) ¹³/₄

Ans: B) ³/₁₁
Solution-:
• Anamika paid ²/₃ as much as Vinita paid
Anamika and Vinita
     2         and      3
Vinita paid ¹/₂ as much as Lalita paid.
Vinita and Lalita
     1     and    2
Racio of Anamika ,Vinita & Lalita is 2:3:6 respectively
.°. fraction of the bill Vinita will pay ³/₁₁ (answer).

53.Rahul invested 20% more than Bikash. Bikash invested 20% less than Kamal. If the total sum of their investment is Rs. 8,280, how much amount did Kamal invest ?

A) 3000

B) 2750

C) 2500

D) 2250

Ans: A) 3000
Solution-:
• Rahul invested 20% more than Bikash.
Rahul and Bikash
     6   and     5
Bikash invested 20% less than Kamal.
Bikash and Kamal
     4     and    5
Racio of Rahul,Bikash & Kamal is 24:20:25 respectively
.°. 69 = 8280 ⇒ 1 = 8280 ÷ 69 = 120
⇒Kamal invest = 25 × 120 = 3000 (answer).

54.Sum the series 40+39+38+37+36+……..+11.

A) 820

B) 775

C) 780

D) 765

Ans: D) 765
Solution-:
• Sum of series 1+2+3+4+5+……+40 = ^{40 × 41}/₂ = 820
Sum of series 1+2+3+4+5+….+10 = ^10×11/₂ = 55
820 – 55 = 765 (answer). OR
n = 30
Use formula S_n = ^{n(first term + last term)}/₂

55.What is the smallest number by which 10976 must be divided to make it a perfect cube ?

A) 2

B) 7

C) 8

D) 4

Ans: D) 4
Solution-:

56.If the sum of two numbers is 42 and the HCF and LCM of these numbers are 6 and 72 respectively, then what is the sum of the reciprocals of the numbers?

A) ⁷/₇₂

B) ⁷²/₇

C) ⁴²/₄₃₃

D) ⁴³³/₄₂

Ans: A) ⁷/₇₂
Solution-:
HCF = 6, LCM = 72
Two numbers are 6a + 6b = 42 (Given)
a + b = 7
ବର୍ତମାନ a ଓ b ର ମାନ 3 ଏବଂ 4 ହେଲେ
ସଙ୍ଖ୍ୟା ଦୁଇଟି 6×3 = 18 ଏବଂ 6×4 = 24 ।
The sum of the reciprocals of the numbers is ¹/₁₈ + ¹ /₂₄ = ⁷/₇₂

57. Rita read ³/₈^thof the book on Monday and ⁴/₅^th of remaining on Tuesday. If there were 30 pages unread, how many pages does the book contains.

A) 120

B) 240

C) 340

D) 230

Ans: B) 240
Solution-:
Rita read ³/₈^thof the book on Monday.So remaining pages = ⁵/₈^th pages.
Tuesday read ⁴/₅^th of the remaining pages.
⁴/₅ × ⁵/₈= ¹/₂
Remaining pages = ⁵/₈ – ¹/₂= ¹/₈ pages.
OR
⁵/₈ × ¹/₅ = ¹/₈
¹/₈ = 30 Pages unread (given).
Book contains 30 × 8 = 240 pages.

58. A number when divided by 296 gives a remainder 75. When the same number is divided by 37, the remainder will be

A) 1

B) 0

C) 5

D) 2

Ans: A) 1
Solution-:

Let the number be N and Quotient be Q.

59. ଏକ ମିଶ୍ରଣରେ କ୍ଷୀର ଓ ଜଳର ଅନୁପାତ 3 : 2 । ଯଦି 4 ଲିଟର ଜଳମିଶା ଯାଏ, ତେବେ କ୍ଷୀର ଓ ଜଳର ପରିମାଣ ସମାନ ହେବ, ତେବେ ଜଳର ପରିମାଣ କେତେ ?

A) 10

B) 15

C) 8

D) 20

Ans: C) 8
Solution-:

  କ୍ଷୀର      ଜଳ
      3  :  2
     3  :   3 (କ୍ଷୀର ଓ ଜଳର ଅନୁପାତକୁ ସମାନ କରାଗଲା)
1 = 4 litre ଜଳ
2= 2 × 4 = 8 litres

60. ଏକ ଘଣ୍ଟାର ଲିଖିତ ମୂଲ୍ୟ 900 ଟଙ୍କା । ଦୋକାନୀ ଏହି ଘଣ୍ଟାକୁ ଦୁଇଟି କ୍ରମାଗତ ରିହାତି ଯଥାକ୍ରମେ 20% ଓ 10% ଦେଇ ବିକ୍ରିକଲା । ଏହାର ବିକ୍ରିମୂଲ୍ୟ କେତେ ହେବ ?

A) ୫୪୮

B) ୬୪୮

C) ୭୪୮

D) ୯୦୦

Ans: B) ୬୪୮
Solution-:

୯୦୦ × ^୮୦/_୧୦୦ × ^୯୦/_୧୦୦ = ୬୪୮
କିମ୍ବା
Successive Discount formula
= x + y – ^xy/₁₀₀
10 + 20 – ²⁰⁰/₁₀₀ = 28%
900 × ²⁸/₁₀₀ = 252 (Discount)
900 – 252 = 648

61. If x²-4x+1 = 0,then what is the value of x³ + ¹/_x³ = ?

A) 48

B) 52

C) 64

D) 44

Ans:B) 52
Solution-: ^x²/_x – ^4x/_x + ¹/_x = ⁰/_x (ଉଭୟ ପାର୍ଶ୍ଵକୁ x ଦ୍ୱାରା ଭାଗକରାଗଲା।)
= x – 4 + ¹/_x = 0
x + ¹/_x = 4
Rule : if
x + ¹/_x = p then x² + ¹/_x² = p²- 2
ଏବଂ x³ + ¹/_x³ = p³- 3p
.°. x³ + ¹/_x³ = 4³ – 3×4
64 – 12 = 52

62. 1f the position of the last 2 digits of a 3 digit number are inter-changed, the new number thus created is 54 higher than the original number. What is the difference between the last two digits of the number?

A) 6

B) 9

C) 12

D) None of these

Ans:A) 6
Solution-:
Always ^Difference/₉ = Answer

63. The simple interest of some principal is its 16/25. If the rate of interest and the time are same, what is the rate of interest?

A) 4

B) 9

C) 8

D) 7

Ans:C) 8
Solution-:
Let the principal be 100.
I = 100 × ¹⁶/₂₅ = 64 (Simple Interest)
t = r (Given)
64 = ^{100 × r²}/₁₀₀
r² = 64 ହେଲେ r = ✓64 = 8

64. ଯଦି କୌଣସି ସଖ୍ୟାର 11 ¹/₉ % ନିଜ ସହିତ ଯୋଡି ଦିଆଯାଏ ତେବେ ଫଳାଫଳ 1000 ହୁଏ , ତେବେ ମୂଳ ସଙ୍ଖ୍ୟା ବାହାର କରନ୍ତୁ ?

A) 924

B) 900

C) 903

D) 824

Ans:B) 900
Solution :- 11 ¹/₉ % = ¹/₉
ଏଠାରେ ମୂଳ ସଂଖ୍ୟା (Original number) ହେଉଛି 9 । ସଂଖ୍ୟାଟିକୁ ନିଜ ସହିତ ଯୋଡି ଦିଆଯାଏ ତେବେ ଫଳାଫଳ 1000 ହୁଏ। ଏହାର ଅର୍ଥ 9 + 1 = 10 = 1000
ହୁଏ ତେବେ ମୂଳ ସଂଖ୍ୟାଟି ହେବ 9 × 100 = 900 କିମ୍ବା
Number + ^Number/₉ = 1000

65. Ram multiplied a number by ³/₅ instead of ⁵/₃ .Find the percentage change in his result?

A) 54%

B) 44%

C) 64%

D) 24%

Ans:C) 64%
Solution :- Let the number be 15 (LCM of 3 & 5)
15 × ³/₅ = 9 (Wrong number)
15 × ⁵/₃= 25 (Right Number)
.°. Riht Number – Wrong Number
25 – 9 = 16
¹⁶/₂₅ × 100 = 64%

66. If a student scores 30% marks then he is failed by 50 marks, but when he scores 50% marks then he is passed by 30 marks. Find the passing %?

A) 33.33%

B) 42.5%

C) 37.5%

D) 40%

Ans:B) 42.5%
Solution:-
Let total mark be x.
ଆମେ ଜାଣିଛନ୍ତି ପାସ ମାର୍କ Constant ରହିବ।
30% + 50 = 50% – 30

^30x/₁₀₀ + 50 = ^50x/₁₀₀ – 30
^20x/₁₀₀ = 80
x = 400 (Total Mark)
Pass mark = 400 × 50% – 30 = 170
¹⁷⁰/₄₀₀ × 100 = 42.5%

67. A dishonest shopkeeper promises to sell his goods at its CP but he uses 750gm weight instead of 1kg. Find the profit % ?

A) 33.33%

B) 42.5%

C) 37.5%

D) 40%

Ans:A) 33.33%
Solution-:
He uses 750gm weight instead of 1kg.
Let 1000g ——-> Rs.1000.
750g —–> Rs.1000 (given)
750g ବିକ୍ରି କଲେ Rs.250 ଟଙ୍କା ଲାଭ ପାଏ।
Profit Rs.250
²⁵⁰/₇₅₀ × 100 = 33.33%

68. A dishonest shopkeeper promises to sell his goods at 25% profit but he uses 40% less weight in 1 kg. find the actual loss% or Profit% ?

A) 33.33%

B) 42.5%

C) 108.33%

D) 40%

Ans:C) 108.33%

Dishonest shopkeeper uses 40% less weight in 1 kg.
Let 1000g ——-> Rs.1000.(CP)
Shopkeeper sells his goods at 25% profit.
ହେଲେ ବିକ୍ରୟମୂଲ୍ୟ = 1000 × ⁵/₄ = 1250 (SP)
600g ବିକ୍ରି କଲେ Rs.650 ଟଙ୍କା ଲାଭ ପାଏ।
Profit℅ = ⁶⁵⁰/₆₀₀ × 100 = 108.33%

69. ଗୋଟିଏ ସମଘନାକାର ପାଣି ଟାଙ୍କିର ଗଭୀରତା 200 ସେ.ମି । ଏଥୁରୁ ଦୈନିକ 1000 ଲିଟର ପାଣି କାଢିଦେଲେ, କେତେ ଦିନରେ ପାଣିତକ ଶେଷ ହୋଇଯିବ ।

A) ୬ ଦିନ

B) ୮ ଦିନ

C) ୪ ଦିନ

D) ୫ ଦିନ

Ans:B) ୮ ଦିନ
ସମାଧାନ -:
ଗୋଟିଏ ସମଘନାକାର ପାଣି ଟାଙ୍କିର ଗଭୀରତା 200 ସେ.ମି ।
ପାଣି ଟାଙ୍କିର ଆୟତନ (Volume) = a³ = 200cm³= 8,000,000 cm³
ମନେରଖ ୧ ଲିଟର = ୧୦୦୦cm³
ହେଲେ 8,000,000 cm³ ÷ 1000cm³ = 8000 ଲିଟର
ପ୍ରତିଦିନ 1000 ଲିଟର ପାଣି କଢାଯାଏ ତେବେ ୮୦୦୦ ଲିଟର ପାଣି କଢାଯିବ ୮ ଦିନରେ । (ଉତ୍ତର)

70. ଏକ ବିଦ୍ୟାଳୟର ଛାତ୍ରମାନଙ୍କୁ ବର୍ଗାକାର କ୍ଷେତ୍ରରେ ଧାଡି ଧାଡି କରି ଠିଆ କରାଇବାରୁ 10 ରୁ କମ ଛାତ୍ର ବଳିପଡିଲେ । ସ୍କୁଲ ଛାତ୍ରସଂଖ୍ୟା 1230 ଜଣ ହେଲେ ପ୍ରତିଧାଡିରେ କେତେ ଜଣ ଛାତ୍ର ଛିଡା ହେଲେ?

A) ୩୩ ଜଣ

B) ୩୦ ଜଣ

C) ୩୫ ଜଣ

D) ୩୮ ଜଣ

Ans:C) ୩୫ ଜଣ
ସମାଧାନ -:
ଯେତେବେଳେ ଛାତ୍ରମାନେ ଏକ ବର୍ଗ ଗଠନ ପାଇଁ ଧାଡିରେ ଛିଡା ହୁଅନ୍ତି, ସେତେବେଳେ ୧୦ରୁ କମ୍ ଛାତ୍ର ବାକି ରହିଥାନ୍ତି ।
୧୨୩୦ର ବର୍ଗମୂଳ ବାହାର କରନ୍ତୁ।
√୧୨୩୦ = ୩୫ ଜଣ, ୫ ଜଣ ବଳକା ରହିବେ। (ଉତ୍ତର)

71. What smallest number has to be deducted from 1294 to leave a reminder of 6 in each case when divided by 9, 11 & 13 ?

A) 1

B) 2

C) 3

D) 4

Ans:A) 1
ସମାଧାନ -: LCM of 9,11,13 = 1287
To leave a reminder of 6 in each case when divided by 9, 11 & 13 we should add 6 to 1287 = 1293
The smallest number 1 has to be deducted from 1294 to leave a reminder of 6 in each case when divided by 9, 11 & 13 (Ans.)

72. Find the smallest number which when divided by 2, 3, 4, 5 & 6 leaves remainder as 1 in each case & 0 when divided by 7?

A) 121

B) 301

C) 202

D) 403

Ans:B) 301
ସମାଧାନ -: LCM of 2,3,4,5,6 = 60
Number (60k + 1) Completely divided by 7.
Let’s check it
60 × 1 + 1 = 61 (not divisible by 7)
60 × 2 + 1 = 121 (not divisible by 7)
60 × 3 + 1 = 181 (not divisible by 7)
60 × 4 + 1 = 241(not divisible by 7)
60 × 5 + 1 = 301 (This number is completely divisible by 7)

73. ଦୁଇଟି ସଂଖ୍ୟାର ଯୋଗଫଳ 15 ଓ ସେମାନଙ୍କର ବର୍ଗର ଯୋଗଫଳ 113 । ତେବେ ସଂଖ୍ୟା ଦ୍ବୟ କେତେ ?

A) 4,11

B) 5,10

C) 7,8

D) 6,9

Ans:C) 7,8
ସମାଧାନ -: a + b = 15
a² + b² = 113
(a+b)² = a² + b² + 2ab
15² = 113 + 2ab
225 = 113 + 2ab
2ab = 225 – 113 = 112
ab = 112 ÷ 2 = 56 ହେଲେ a,b = 7,8 ହେବ।

74. Pointing to a boy Madhu says, “he is the only son of my father-in-law’s only son”. How that lady is related to that boy ?

A) Husband

B) Wife

C) Mother

D) Son

Ans:C) Mother
ସମାଧାନ -: Tree diagram

75. The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 25% respectively. Find the percentage change in the volume of the cuboid.

A) 35%

B) 75%

C) 65%

D) 25%

Ans:C) 65%
ସମାଧାନ -:

76. In an competitive examination, 66% candidates passed in section A and 51% in section B, while 8% failed in both the sections. If 450 students passed in both the sections, find the total number of candidates who appeared in the competitive examination.

A) 1900

B) 1800

C) 2000

D) 1700

Ans:B) 1800
ସମାଧାନ -: 66% Candidates passed in section A.
51% Candidates passed in section B.
Total candidates passed in both section =117%
8% Candidates failed in both the section it means 92% of students passed in both the section.
117% – 92% = 25% = 450 (Given)
100% = 1800

77. In an election between two candidates, the candidate who gets 30% of the total votes polled is defeated by 16000 votes. What is the total number of votes polled ?

A) 20000

B) 24000

C) 28000

D) 40000

Ans:D) 40,000
ସମାଧାନ -: The candidate who gets 30% of the total votes polled is defeated by 16000 votes.
40% = 16000
100% = 16000 × 2.5 = 40,000

78. 200 ଲୋକଙ୍କ ମଧ୍ୟରେ 140 ଜଣ ଚାହା ପସନ୍ଦ କରନ୍ତି, 120 ଜଣ କଫି ପସନ୍ଦ କରନ୍ତି ଓ 80 ଜଣ ଉଭୟ ଚାହା ଓ କଫି ପସନ୍ଦ କରନ୍ତି ।ତେବେ କେତେଜଣ ଚାହା ବା କଫି କୌଣସିଟି ପସନ୍ଦ କରନ୍ତି ନାହିଁ ?

A) 76

B) 58

C) 48

D) 20

Ans:D) 20
ସମାଧାନ -: Alternate Method :-
140 ଜଣ ଚାହା ପସନ୍ଦ କରନ୍ତି ।
120 ଜଣ କଫି ପସନ୍ଦ କରନ୍ତି ।
ଚାହା ଓ କଫି ପସନ୍ଦ କରୁଥିବା ଲୋକସଂଖ୍ୟା = ୧୨୦ + ୧୪୦ = ୨୬୦
260 ଜଣଙ୍କ ମଧ୍ୟରୁ 80 ଜଣ ଉଭୟ ଚାହା ଓ କଫି ପସନ୍ଦ କରନ୍ତି । ତେଣୁ ୨୬୦ – ୮୦ = ୧୮୦ ଜଣ ଊଭୟ ଚାହା ଓ କଫି ପସନ୍ଦ କରନ୍ତି।
.°. 200 – 180 = 20 ଜଣ ଚାହା ବା କଫି କୌଣସିଟି ପସନ୍ଦ କରନ୍ତି ନାହିଁ।

79. Sum of two numbers is 25 and their difference is 13. Find their product.
ଦୁଇଟି ସଂଖ୍ୟାର ଯୋଗଫଳ 25 ଓ ବିୟୋଗଫଳ 13 ଅଟେ ।ସଂଖ୍ୟା ଦୁଇଟିର ଗୁଣଫଳ କେତେ ?

A) 196

B) 256

C) 114

D) 169

Ans:C) 114
ସମାଧାନ -: (a + b)² = (a – b)² + 4ab
OR
ଦୁଇଟି ସଂଖ୍ୟାର ଯୋଗଫଳ 25 ଓ ବିୟୋଗଫଳ 13 ଅଟେ ।
a + b = 25
a – b = 13
ଯୋଗ କଲେ ପାଇବା 2a = 38
a = 19 ହେଲେ b = 6
.°. ab = 19 × 6 = 114

80. 20 ଜଣ ପୁରୁଷ ଓ 25 ଜଣ ସ୍ତ୍ରୀ ଗୋଟିଏ କାର୍ଯ୍ୟକୁ 5 ଦିନରେ ଶେଷ କରନ୍ତି । ଜଣେ ପୁରୁଷକୁ ଏକାକୀ ସମାନ କାର୍ଯ୍ୟ ଶେଷ କରିବାକୁ 150 ଦିନ ଲାଗେ । ତେବେ ଜଣେ ସ୍ତ୍ରୀକୁ ଏକାକୀ ସମାନ କାର୍ଯ୍ୟ ଶେଷ କରିବାକୁ କେତେ ଦିନ ଲାଗିବ ?

A) 300

B) 375

C) 250

D) 348

Ans:B) 375
ସମାଧାନ -: 20 ଜଣ ପୁରୁଷ ଓ 25 ଜଣ ସ୍ତ୍ରୀ ଗୋଟିଏ କାର୍ଯ୍ୟକୁ 5 ଦିନରେ ଶେଷ କରନ୍ତି ।ଜଣେ ପୁରୁଷକୁ ଏକାକୀ ସମାନ କାର୍ଯ୍ୟ ଶେଷ କରିବାକୁ 150 ଦିନ ଲାଗେ ।
(20 ପୁରୁଷ + 25 ସ୍ତ୍ରୀ) × 5 ଦିନ = 1 ପୁରୁଷ × 150 ଦିନ (ସମାନ କାର୍ଯ୍ୟ ପାଇଁ)
100 ପୁରୁଷ + 125 ସ୍ତ୍ରୀ = 150 ପୁରୁଷ
125 ସ୍ତ୍ରୀ = 50 ପୁରୁଷ
5 ସ୍ତ୍ରୀ = 2 ପୁରୁଷ
ଜଣେ ପୁରୁଷକୁ ଏକାକୀ ସମାନ କାର୍ଯ୍ୟ ଶେଷ କରିବାକୁ 150 ଦିନ ଲାଗେ । 2 ଜଣ ପୁରୁଷ ସେହି କାର୍ଯ୍ୟକୁ 75 ଦିନରେ ସାରିବେ। 5 ଜଣ ସ୍ତ୍ରୀ ସେହି କାର୍ଯ୍ୟକୁ 75 ଦିନରେ ସାରିବେ। .°.1 ଜଣ ସ୍ତ୍ରୀ ସେହି କାର୍ଯ୍ୟକୁ 75 × 5 = 375 ଦିନରେ ସାରିବେ।

81. ଯଦି ଗୋଟିଏ ସମବାହୁ ତ୍ରିଭୂଜର ପ୍ରତ୍ୟେକ ବାହୁର ଦୈର୍ଘ୍ୟକୁ 2 ସେ.ମି. କମାଯାଏ, ଏହାର କ୍ଷେତ୍ରଫଳ 11✓3 ବର୍ଗ ସେ.ମି. କମିଯାଏ । ତ୍ରିଭୂଜଟିର ଉଚ୍ଚତା କେତେ ?

A) 3√3 cm

B) 5√3 cm

C) 6√3 cm

D) 4√3 cm

Ans:C) 6√3
ସମାଧାନ -: ^✓3/₄a² – ^✓3/₄(a-2)² = 11√3
^✓3/₄{a² – (a² – 2.a.2 + 2²)} = 11√3
^{a² – a² + 4a – 4}/₄ = 11
4a – 4 = 44
4(a – 1) = 44
a – 1 = 11 ହେଲେ a = 12 (ସମବାହୁ∆ ର ବାହୁ)
ସମବାହୁ∆ ର ଉଚ୍ଚତା = ^√3/₂ × a

82. A person deposited Rs. 9000 in a bank. After 2 years, he withdrew Rs. 4000. At the end of 5 years he got Rs. 7640 from the bank. Find out the rate of simple interest per annum ?
ଜଣେ ବ୍ୟକ୍ତି ଏକ ବ୍ୟାଙ୍କ୍‌ରେ 9000 ଟଙ୍କା ଜମା ଦେଲା । 2 ବର୍ଷ ପରେ ସେ 4000 ଟଙ୍କା ଉଠାଇଲା । 5 ବର୍ଷ ଶେଷରେ ସେ ବ୍ୟାଙ୍କ୍‌ରୁ 7640 ଟଙ୍କା ପାଇଲା । ତେବେ ବାର୍ଷିକ ସରଳ ସୁଧର ହାର କେତେ ?

A) 6%

B) 10%

C) 8%

D) 12%

Ans:C) 8%
ସମାଧାନ -: (^{9000 × 2 × r}/₁₀₀) +
(^{5000 × 3 × r}/₁₀₀) = 2640 (Int.)

83. What will be the total compound interest for 9 months on Rs. 5,000 at the rate of 40% per annum if the interest is compounded quarterly?

5000 ଟଙ୍କାରେ ବାର୍ଷିକ ଶତକଡା 40 ଚକ୍ରବୃଦ୍ଧି ସୁଧ ହାରରେ 9 ମାସର ମୋଟ ଚକ୍ରବୃଦ୍ଧି ସୁଧ କେତେ ହେବ. ଯଦି 3 ମାସକୁ ଥରେ ଚକ୍ରବୃଦ୍ଧି ସୁଧ ହିସାବ କରାଯାଏ ?

A) 1655

B) 1645

C) 1665

D) 1630

Ans:A) 1655
ସମାଧାନ -: P = 5000 , T = ⁹/₃ = 3, r = ⁴⁰/₄ = 10%
A = p × ( 1 + ^r/₁₀₀)^t
Compound Interest = A – P
6655 – 5000 = 1655

84. ¹/_5×6 + ¹/_6×7 + ¹/_7×8 + ¹/_{8×9………}16 ଟି ପଦ ପର୍ଯ୍ୟନ୍ତ ଅନୁକ୍ରମର ଯୋଗଫଳ କେତେ ?

A) ¹⁶/₁₀₅

B) ⅘

C) ²⁰/₂₁

D) ²¹/₂₂

Ans:A) ¹⁶/₁₀₅
ସମାଧାନ -: Formula -: ¹/_n(n+1)= ¹/_n – ¹/_n+1
¹/₅ – ¹/₂₁ = ¹⁶/₁₀₅

85. The product of two numbers is 70 and their addition is 17. What is the difference between the two ?

A) 5

B) 6

C) 3

D) 4

Ans:C) 3
ସମାଧାନ -: a + b = 17
b = 17 – a
a × (17 – a) = 70
17a – a² = 70
a² – 17a = -70 (ଉଭୟ ପାର୍ଶ୍ଵକୁ – ଦ୍ଵାରା ଗୁଣନ କରାଗଲା)
a² – 17a + 70 = 0
a² + (-10-7)a + 70 = 0
a² -10a-7a + 70 = 0
a(a-10)-7(a – 10)
(a – 7) (a-10)
a = 7 & 10
.°. Difference between the two number is 10 – 7 = 3

86. ରହିମ୍ କିଛି କୁକୁଡା ଓ ଛେଳି ପାଳିଥ୍‌ଲା । ଯଦି ଛେଳି ଓ କୁକୁଡା ମୁଣ୍ଡ ମିଶି ୯୦ ଓ ଗୋଡ ମିଶି ୨୪୮ ହୁଏ, ତେବେ ରହିମ୍ କେତୋଟି ଛେଳି ପାଳିଥ୍‌ଲା ?

A) ୩୨

B) ୩୬

C) ୩୪

D) ୪୦

Ans:C) ୩୪
ସମାଧାନ -: H + G = 90 (ମୋଟ ମୁଣ୍ଡ ସଂଖ୍ୟା)
2H + 4G = 248 (ଗୋଡର ମୋଟ ସଂଖ୍ୟା)
2H + 4G = 248
– (2H + 2G = 180)
2G = 68 ହେଲେ। G = 34

87. The average of 6 members in family is 22 years. If the age of the youngest is 7 years, what was average age of the members of the family at the time of birth of the youngest?

A) 15

B) 17

C) 18

D) 16

Ans:C) 18
Solution -: 22 × 6 = 132 (୬ ଜଣଙ୍କ ମୋଟ ବୟସ)
Age of the members of the family at the time of birth of the youngest was 132 – 6 × 7 = 90
Average age of 5 member was 90 ÷ 5 = 18

88. If the side of a square is increased by 20%, the area of the square is increased by:

A) 40%

B) 35%

C) 44%

D) 25%

Ans:C) 44%
Solution -: Increase Percentage formula = x + y + ^xy/₁₀₀
20 + 20 + ⁴⁰⁰/_{100 = 44%}

89. If the area of a circle is decreased by 36% then radius of a circle decreases by

A) 10%

B) 35%

C) 24%

D) 20%

Ans:D) 20%
Solution -:
Initial radius be “r”.
Area = πr²
The area of a circle is decreased by 36% = πr² × ⁶⁴/_{100 = 0.64πr² ⮕ π(0.8)² (ଏମିତି ଲେଖିଲେ ‘r’ ର ମାନ ପାଇହେବ)
Radius decreased = r – 0.8r = 0.2r ⮕ Decreased % = ^0.2r/_r × 100 = 20%}
Alternative Method-:

90. The product of two numbers is 130 and their addition is 23. What is the difference between the two?

A) 5

B) 3

C) 7

D) 10

Ans:B) 3
Solution:-
a × b = 130
a + b = 23
formula -: (a-b)² = (a + b)² – 4ab
23² – 4 × 130
529 – 520 = 09
a – b = √9 = 3 (Ans)

91. The radius of a circle is increased by 1%. The percentage increase in area is:

A) 1%

B) 1.1%

C) 2%

D) 2.01%

Ans:D) 2.01%
Solution:-
Successive increase formula: x + y + ^xy/₁₀₀
1+1 + ^1×1/₁₀₀ = 2.01%

92.

A) 13

B) 1

C) √13

D) 13¹⁵/₁₆

Ans:D) 13¹⁵/₁₆
Solution:-

93.What is the rate of interest if a sum of Rs.4000/- becomes Rs.5000/- in 5 years on simple interest?

A) 5%

B) 10%

C) 20%

D) 13%

Ans:A) 5%
Solution:-
p = 4000, t = 5 years, A = 5000 ହେଲେ ସୁଧ(i) = 1000
4000 ଟଙ୍କାର 5 ବର୍ଷରେ ସୁଧ 1000 ହେଲେ, 4000 ହଜାର ଟଙ୍କାର ବର୍ଷକ ସୁଧ ହେବ 200 ଟଙ୍କା।
ସୁଧର ହାର = ²⁰⁰/_{4000 × 100 = ¹/₂₀ = 5%}

94.Reeta says that she sells her goods at the cost price. If she uses 80 gram weight instead of 100 gram, then what is her profit percentage?

A) 25%

B) 20%

C) 40%

D) 5%

Ans:A) 25%

If she uses 80 gram weight instead of 100 gram. It means direct 20 gram profit when he sell 80 grams goods.
profit % = ²⁰/₈₀ × 100 = 25%

95. 49¹⁵ – 1 is completely divisible by which number?

A) 8

B) 7

C) 50

D) 29

Ans: A) 8

a^m – b^m/_a-b is completely divisible for all m.
a^m – b^m/_a+b is completely divisible if m is even.
a^m + b^m/_a+b is completely divisible if m is odd.
Solution-: 49¹⁵ – 1 କୁ ଆମେ ଲେଖିପାରିବା 49¹⁵ – 1¹⁵
Put the formula and get the answer.
a^m – b^m/_a-b is completely divisible for all m.
^{49¹⁵ – 1¹⁵}/_{49 – 1}
.°. 49¹⁵ – 1¹⁵ ସଂଖ୍ୟାଟି 49 – 1 ବା 48 ବା ତାହାର ଗୁଣିତକ(Factors) ଦ୍ଵାରା ଭାଗ ହେବ।

96. The day before yesterday was Saturday. What day will it be the day after tomorrow?

A) Monday

B) Wednesday

C) Thursday

D) Friday

Ans: B) Wednesday

Solution-:

97. The Two number are in the ratio 3: 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:

A) 28

B) 55

C) 33

D) 64

Ans: C) 33

Solution-:ବର୍ତମାନ ଆମେ ଅନୁପାତର ଅନ୍ତରକୁ ସମାନ କରିବା।
(3:5) × 11
(12:23) × 2
33:55
24:46

98. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A) 276

B) 279

C) 322

D) 385

Ans: C) 322

Solution-:Let two numbers be 23a,23b.
H.C.F. of two numbers is 23.(Given)
The other two factors of their L.C.M. are 13 and 14.(Given)
The L.C.M. of 23a,23b is 23ab .
Factors of 23ab = 23 × a × b
.°. The two numbers are 23 × 13 & 23 × 14 (23a,23b).
Largest number is 23 × 14 = 322

99. A person takes a loan of Rs. 2500 at 4% simple interest. He returns Rs.1500 at the end of 1 year. What amount he would pay to clear his dues at the end of 2 years?

A) 1150

B) 1050

C) 1140

D) 1100

Ans: C) 1140

Solution-:A person takes a loan of Rs. 2500 at 4% simple interest.
p = 2500 r = 4% i = 100———> 1st. year
p = 1000 r = 4% i = 40 ——> 2nd. year
Amount he would pay to clear his dues at the end of 2 years ia 1000 + 100 + 40 = 1140

100. If a number is increased by 50% and then decreased by 60%, the resulting number becomes 80 less than the original number. What was the original number?

A) 80

B) 130

C) 200

D) 100

Ans: C) 200

Solution-: Let the number be N is increased by 50% and then decreased by 60%.
N—–50%↑—^3N/₂
^3N/₂—-60%↓—^6N/₁₀
The resulting number becomes 80 less than the original number.
N – 80 = ^6N/₁₀
10N – 800 = 6N
4N = 800 => N = 200

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