Arithmetic Reasoning — Part 2 MCQs with Explanations

'All but nine died' means 'All except nine died' i.e. 9 sheep remained alive.

10 is the correct answer to “In a family, the father took 1/4 of the cake and he had 3 times as much as each of the other members had. The total number of family members is”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Let R, G and B represent the number of balls in red, green and blue boxes respectively. Then, . R + G + B = 108 ...(i), G + R = 2B ...(ii) B = 2R ...(iii) From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R. Putting G = 3R and B = 2R in (i), we get: R + 3R + 2R = 108 ⇔ 6R = 108 ⇔ R = 18. Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

Total runs scored = (36 x 5) = 180. Let the runs scored by E be x. Then, runs scored by D = x + 5; runs scored by A = x + 8; runs scored by B = x + x + 5 = 2x + 5; runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x. So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120. Therefore 3x + 120 =180 ⇔ 3X = 60 ⇔ x = 20.

Total number of digits = (No. of digits in 1- digit page nos. + No. of digits in 2-digit page nos. + No. of digits in 3- digit page nos.) = (1 x 9 + 2 x 90 + 3 x 267) = (9 + 180 + 801) = 990.

Let d and s represent the number of daughters and sons respectively. Then, we have : d - 1 = s and 2 (s - 1) = d. Solving these two equations, we get: d = 4, s = 3.

60 is the correct answer to “At a dinner party every two guests used a bowl of rice between them, every three guests used a bowl of dal between them and every four used a bowl of meat between them. There were altogether 65 dishes. How many guests were present at the party”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Ayush's present age = 10 years. His mother's present age = (10 + 20) years = 30 years. Ayush's father's present age = (30 + 5) years = 35 years. Ayush's father's age at the time of Ayush's birth = (35 - 10) years = 25 years. Therefore Ayush's father's age at the time of marriage = (25 - 2) years = 23 years.

Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25.

None of these is the correct answer to “If a clock takes seven seconds to strike seven, how long will it take to strike ten”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Let the number of cows be x and the number of hens be y. Then, 4x + 2y = 2 (x + y) + 14 ⇔ 4x + 2y = 2x + 2y + 14 ⇔ 2x = 14 ⇔ x = 7.

Let the father's age be x and the son's age be y. Then, x - y = y or x = 2y, Now, x = 36. So, 2y = 36 or y = 18. Therefore Son's present age = 18 years. So, son's age 5 years ago = 13 years.

23/30 is the correct answer to “In a class, 3/5 of the students are girls and rest are boys. If 2/9 of the girls and 1/4 of the boys are absent, what part of the total number of students is present”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Let the daughter's age be x years. Then, father's age = (3x) years. Mother's age = (3x - 9) years; Son's age = (x + 7) years. So, x + 7 = (3x-9)/2 ⇔ 2x + 14 = 3x - 9 ⇔ x = 23. Therefore Mother's age = (3X - 9) = (69 - 9) years = 60 years.

1 billion km is the correct answer to “If a 1 mm thick paper is folded so that the area is halved at every fold, then what would be the thickness of the pile after 50 folds”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

15 is the correct answer to “First bunch of bananas has (1/4) again as many bananas as a second bunch. If the second bunch has 3 bananas less than the first bunch, then the number of bananas in the first bunch is”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Numbers from 1 to 60, which are divisible by 6 are : 6,12,18, 24, 30, 36,42, 48, 54, 60. There are 10 such numbers. Numbers from 1 to 60, the sum of whose digits is 6 are : 6, 15, 24, 33, 42, 51, 60. There are 7 such numbers of which 4 are common to the above ones. So, there are 3such uncommon numbers. Numbers from 1 to 60, which have 6 as one of the digits are 6, 16, 26, 36, 46, 56, 60. Clearly, there are 4 such uncommon numbers. So, numbers 'not connected with 6' = 60 - (10 + 3 + 4) = 43.

Let the number be x. Then, x + 13x = 112 ⇔ 14x = 112 ⇔ x = 8.

The seven pieces consist of 6 smaller equal pieces and one half cake piece. Weight of each small piece = 20 g. So, total weight of the cake = [2 x (20 x6)]g= 240 g.

Let the number of 20-paise coins be x. Then, number of 25-paise coins = (324 - x). Therefore 0.20 x x + 0.25 (324 - x) = 71 ⇔ 20x + 25 (324 - x) = 7100 ⇔ 5x= 1000 ⇔ x = 200. Hence, number of 25-paise coins = (324 - x) - 124.

Clearly, the black cards are either clubs or spades while the red cards are either diamonds or hearts. Let the number of spades be x. Then, number of clubs = (7 - x). Number of diamonds = 2 x number of spades = 2x; Number of hearts = 2 x number of diamonds = 4x. Total number of cards = x + 2x + 4x + 7 - x = 6x + 7. Therefore 6x + 7 = 13 ⇔ 6x = 6 ⇔ x - 1. Hence, number of clubs = (7 - x) = 6.

Let the fixed charge be Rs. x and variable charge be Rs.y per km. Then, x + 16y = 156 ...(i) and x + 24y = 204 ...(ii) Solving (i) and (ii), we get: x = 60, y = 6. Therefore Cost of travelling 30 km = 60 + 30 y = Rs. (60 + 30 x 6) = Rs. 240.

133 is the correct answer to “If readymade shirts need alterations 2 out of 3 in sleeves, 3 out of 4 in collar and 4 out of 5 in the body, how many alterations will be required for 60 shirts”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Clearly, total number of handshakes = (9+ 8 + 7 + 6 + 5 + 4 + 3 + 2+1) = 45.

Let the total number of sweets be (25x + 8). Then, (25x + 8) - 22 is divisible by 28 ⇔ (25x - 14) is divisible by 28 ⇔ 28x - (3x + 14) is divisible by 28 ⇔ (3x + 14) is divisible by 28 ⇔ x = 14. Therefore Total number of sweets = (25 x 14 + 8) = 358.

Clearly, out of every 16 persons, there is one captain. So, number of captains (1200/16) = 75.

Let number of keepers be x. Then, Total number of feet = 2 x 50 + 4 x 45 + 4 x 8 + 2x = 2x + 312. Total number of heads = 50 + 45 + 8 + x= 103 + x. Therefore (2x + 312) = (103 + x) + 224 or x = 15.

Net ascent of the monkey in 1 hour = (30 - 20) feet = 10 feet. So, the monkey ascends 90 feet in 9 hours i.e. till 5 p.m. Clearly, in the next 1 hour i.e. till 6 p.m. the monkey ascends remaining 30 feet to touch the flag.

Let the ten's digit be x. Then, unit's digit = (11 - x). So, number = 10x + (11 - x) = 9x + 11. Therefore (9x + 11) + 27 = 10 (11 - x) + x ⇔ 9x + 38 = 110 - 9x ⇔ 18x = 72 ⇔ x = 4. Thus, ten's digit = 4 and unit's digit = 7. Hence, required number = 47.

Income on the first day = Re. 1. Income on the 2nd day = Rs. (1x2) = Rs. 2 1 . Income on the 3rd day = Rs. (2 1 x 2) = Rs. 2 2 and so on. Thus, Income on the rath day = Rs. 2 n-1 . Therefore Income on the 10th day = Rs. 2 9 .

Clearly, we have to first find two numbers whose difference is 2 and of which the smaller one is a perfect square and the bigger one a perfect cube. Such numbers are 25 and 27. Thus, Nitin is now 26 years old. Since the next perfect cube after 27 is 64, so required time period = (64 - 26) years = 38 years.

Step 1: Let the total number of sweets be = S. Step 2: If all 175 children were present, each child would get = S ÷ 175 sweets. Step 3: But 35 children were absent, so only 140 children were present. Step 4: Now each child got = S ÷ 140 sweets. Step 5: It is given that each child got 4 more sweets than originally planned. So, (S ÷ 140) − (S ÷ 175) = 4 Step 6: Take LCM of denominators: LCM of 140 and 175 = 700. (S × (5 − 4)) ÷ 700 = 4 S ÷ 700 = 4 Step 7: Therefore, S = 700 × 4 = 2800 Final Answer: Total sweets available = 2800

Clearly, number of ways of arranging 5 books = 5 ! = 5 x 4 x 3 x 2 x 1 = 120. So, total time taken = 120 minutes = 2 hours.

Since the number of carbons is 2, only two copies can be obtained.

No. of digits in 1-digit page nos. = 1x9 = 9. No. of digits in 2-digit page nos. = 2 x 90 = 180. No. of digits in 3-digit page nos. = 3 x 900 = 2700. No. of digits in 4-digit page nos. = 3189 - (9 + 180 + 2700) = 3189 - 2889 = 300. Therefore No. of pages with 4-digit page nos. = (300/4) = 75. Hence, total number of pages = (999 + 75) = 1074.