Arithmetic Reasoning — Part 1 MCQs with Explanations

Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A. Then, 2x + 3y = 77 ...(i) and 3x + 2y = 73 ...(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17. Putting y = 17 in (i), we get: x = 13.

Data inadequate

8 is the correct answer to “A number of friends decided to go on a picnic and planned to spend Rs. 96 on eatables. Four of them, however, did not turn up. As a consequence, the remaining ones had to contribute Rs. 4 each extra. The number of those who attended the picnic was”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Clearly, we have : B - 3 = E ...(i) B + 3 = D ...(ii) A+B = D + E+10 ...(iii) B = C + 2 ...(iv) A+B + C + D + E= 133 ...(v) From (i) and (ii), we have : 2 B = D + E ...(vi) From (iii) and (vi), we have : A = B + 10 ...(vii) Using (iv), (vi) and (vii) in (v), we get: (B + 10) + B + (B - 2) + 2B = 133 ⇔ 5B = 125 ⇔ B = 25.

4 is the correct answer to “A pineapple costs Rs. 7 each. A watermelon costs Rs. 5 each. X spends Rs. 38 on these fruits. The number of pineapples purchased is”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Let x and y be the ten's and unit's digits respectively of the numeral denoting the woman's age. Then, woman's age = (10X + y) years; husband's age = (10y + x) years. Therefore (10y + x)- (10X + y) = (1/11) (10y + x + 10x + y) ⇔ (9y-9x) = (1/11)(11y + 11x) = (x + y) ⇔ 10x = 8y ⇔ x = (4/5)y Clearly, y should be a single-digit multiple of 5, which is 5. So, x = 4, y = 5. Hence, woman's age = 10x + y = 45 years.

Clearly, while counting, the numbers associated to the thumb will be : 1, 9,17, 25,..... i.e. numbers of the form (8n + 1). Since 1994 = 249 x 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger.

Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.

Since B and D are twins, so B = D. Now, A = B + 3 and A = C - 3. Thus, B + 3 = C - 3 ⇔ D + 3 = C-3 ⇔ C - D = 6.

Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.

Each row contains 12 plants. There are 11 gapes between the two corner trees (11 x 2) metres and 1 metre on each side is left. Therefore Length = (22 + 2) m = 24 m.

Manick's present age = 12 years, Rahul's present age = 4 years. Let Manick be twice as old as Rahul after x years from now. Then, 12 + x = 2 (4 + x) ⇔ 12 + x = 8 + 2x ⇔ x = 4. Hence, Manick's required age = 12 + x = 16 years.

Number of cuts made to cut a roll into 10 pieces = 9. Therefore Required number of rolls = (45 x 24)/9 = 120.

Let the number of boys and girls participating in sports be 3x and 2x respectively. Then, 3x = 15 or x = 5. So, number of girls participating in sports = 2x = 10. Number of students not participating in sports = 60 - (15 + 10) = 35. Let number of boys not participating in sports be y. Then, number of girls not participating in sports = (35 -y). Therefore (35 - y) = y + 5 ⇔ 2y ⇔ 30 ⇔ y = 15. So, number of girls not participating in sports = (35 - 15) = 20. Hence, total number of girls in the class = (10 + 20) = 30.

Let x and y be the number of deer and peacocks in the zoo respectively. Then, x + y = 80 ...(i) and 4x + 2y = 200 or 2x + y = 100 ...(ii) Solving (i) and (ii), we get) x = 20, y = 60.

Since there are socks of only two colours, so two out of any three socks must always be of the same colour.

Total number of routes from Bristol to Carlisle = (4 x 3 x 2) = 24.

Let money with Ken = x. Then, money with Mac = x + £ 3. Now, 3x = (x + x + £ 3) + £ 2 ⇔ x = £ 5. Therefore Total money with Mac and Ken = 2x + £ 3 = £ 13.

Let the number of boys be x. Then, (3/4)x = 18 or x = 18 x(4/3) = 24. If total number of students is y, then (2/3) y = 24 or y = 24 x (3/2) = 36. Therefore Number of girls in the class = (36 - 24) = 12.

Let son's age be x years. Then, father's age = (3x) years. Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years. So, 3x - 5 = 4 (x - 5) ⇔ 3x - 5 = 4x - 20 ⇔ x = 15.

5/9 is the correct answer to “A waiter's salary consists of his salary and tips. During one week his tips were 5/4 of his salary. What fraction of his income came from tips”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39. So, required number = 10 + 10 = 20.

100 days is the correct answer to “If 100 cats kill 100 mice in 100 days, then 4 cats would kill 4 mice in how many days”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

L.C.M. of 6, 5, 7, 10 and 12 is 420. So, the bells will toll together after every 420 seconds i.e. 7 minutes. Now, 7 x 8 = 56 and 7 x 9 = 63. Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.

Originally, let number of women = x. Then, number of men = 2x. So, in city Y, we have : (2x - 10) = (x + 5) or x - 15. Therefore Total number of passengers in the beginning = (x + 2x) = 3x = 45.

Clearly, we have : A = B - 3 ...(i) D + 5 = E ...(ii) A+C = 2E ...(iii) B + D = A+C = 2E ...(iv) A+B + C + D + E=150 ...(v) From (iii), (iv) and (v), we get: 5E = 150 or E = 30. Putting E = 30 in (ii), we get: D = 25. Putting E = 30 and D = 25 in (iv), we get: B = 35. Putting B = 35 in (i), we get: A = 32. Putting A = 32 and E = 30 in (iii), we get: C = 28.

Since each pole at the corner of the plot is common to its two sides, so we have : Total number of poles needed = 27 x 4 - 4 = 108 - 4 = 104.

75% is the correct answer to “In a city, 40% of the adults are illiterate while 85% of the children are literate. If the ratio of the adults to that of the children is 2 : 3, then what percent of the population is literate”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

We have : A = 3B ...(i) and C - 4 = 2 (A - 4) ...(ii) Also, A + 4 = 31 or A= 31-4 = 27. Putting A = 27 in (i), we get: B = 9. Putting A = 27 in (ii), we get C = 50.

Let Varan's age today = x years. Then, Varan's age after 1 year = (x + 1) years. Therefore x + 1 = 2 (x - 12) ⇔ x + 1 = 2x - 24 ⇔ x = 25.

There were all sparrows but six' means that six birds were not sparrows but only pigeons and ducks. Similarly, number of sparrows + number of ducks = 6 and number of sparrows + number of pigeons = 6. This is possible when there are 3 sparrows, 3 pigeons and 3 ducks i.e. 9 birds in all.

£ 330 is the correct answer to “Mr. Johnson was to earn £ 300 and a free holiday for seven weeks' work. He worked for only 4 weeks and earned £ 30 and a free holiday. What was the value of the holiday”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Clearly, the smallest such number is 3. Three ducks can be arranged as shown above to satisfy all the three given conditions.

3/11 is the correct answer to “Three friends had dinner at a restaurant. When the bill was received, Amita paid 2/3 as much as Veena paid and Veena paid 1/2 as much as Tanya paid. What fraction of the bill did Veena pay”. This question is standard in competitive examination syllabi for Arithmetic Reasoning.

Let total number of members be 100, Then, number of members owning only 2 cars = 20. Number of members owning 3 cars = 40% of 80 = 32. Number of members owning only 1 car = 100 - (20 + 32) = 48. Thus, 48% of the total members own one car each.

When Rahul was born, his brother's age = 6 years; His father's age = (6 + 32) years = 38 years, His mother's age = (38 - 3) years = 35 years; His sister's age = (35 - 25) years = 10 years.

Let number of horses = number of men = x. Then, number of legs = 4x + 2 x (x/2) = 5x. So, 5X = 70 or x = 14.

When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years. Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32. Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years. Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.

Let number of girls = x and number of boys = 3x. Then, 3x + x = 4x = total number of students. Thus, to find exact value of x, the total number of students must be divisible by 4.