Q1. The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?
(a) 21
(b) 25
(c) 41
(d) 67
Sol.
Using options, we find that four consecutive odd numbers are 37, 39, 41 and 43. The sum of these 4 numbers is 160, when divided by 10, we get 16 which is a perfect square.
Therefore, 41 is one of the odd numbers.
Q2. When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?
(a) 5
(b) 6
(c) 7
(d) 8
Sol.
Let the number be (10x+y), so when the digits of number are reversed the number becomes (10y+x). According the question, (10y+x)-(10x+y)=18
9(y-x)=18 y-x=12
So, the possible pairs of (x, y) are
(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9).
But we want the number other than 13. Thus, there are 6 possible numbers ie, 24, 35, 46, 57, 68, 79.
So, total number of possible numbers are 6.
Q3. The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be
(a) 101 : 88
(b) 87 : 100
(c) 110 : 111
(d)97 : 84
Sol.
Using options, we find that sum of numerator and denominator of 97 : 84 is (97+84)=181 which is a prime number. Hence, it is the appropriate answer.
Q4. If Price of sugar having been fallen by 10%. then consumer can buy 21 kg. more than before. Had the price been increased by 16.6% how much quantity of sugar could he have bought for the same sum.
(a) 180 kg
(b) 162kg
(c) 210 kg
(d) None of these
Sol.
Decrease in quantity = 10% = 1/10
Quantity brought is increased by 1/9 of original quantity
Original quantity = (9 × 21) kg
= 189 kg.
If price is increased by 16.66% = 1/6
So, decreased quantity = 1-1/7=6/7
Quantity brought = 9 × 21 × 6/7
= 162 kg.
Q5. What is the remainder when 1044 × 1047 × 1050 × 1053 is divided by 33?
(a) 13
(b) 22
(c) 30
(d) 8
Sol.
Remainder = (1044 × 1047 × 1050 × 1053)/33
=(21×24×27×30)/33
=(21×24×-6×-3)/33( by using remainder theorem)
=(21×24×18)/33=(-12×-9×18)/33
⇒(12×9×18)/33==30/33
Remainder = 30
Q6. Rs. 4000 is divided into two parts such that one part is put out at 3% and the other at 5% rate. If the annual interest earned from both the investments be Rs. 144, find the first part.
(a) Rs. 4000
(b) Rs. 2800
(c) Rs. 3800
(d) Rs. 3500
Sol.
Average rate = 144/4000 × 100 = 3.6%
Ratio = 14:6 = 7:3(using alligation method
So, first part = 7/10 × 4000 = Rs. 2800
Q7. The average age of 24 students and the class teacher is 16 years. If the class teacher’s age is excluded, the average is reduced by one year. Find the age of the class teacher.
(a) 40 years
(b) 38 years
(c) 50 years
(d) 56 years
Sol.
Class teacher’s age = 25 × 16 – 24 × 15
= 400 – 360 = 40 years
Q8. A person saves 15% of his income. If his income increase by 35% and he starts savings 20% of his income, by what percentage does his savings increase?
(a) 60%
(b) 70%
(c) 75%
(d) 80%
Sol.
Assume his initial income = Rs. 100, so savings = 15
New income = 135 ⇒ new savings = 27
Percentage increase in savings = 80%
Q9. An amount doubles itself in 5 years at S.I. In how many years will its interest become its 250%?
(a) 10 years
(b) 15/2 years
(c) 25/2 years
(d) 15 years
Sol.
Amount doubles itself → 100% is added to the principal in five years → 20% is added to principal every year.
So, Rate of interest = 20% For interest to become 250%, time required = 250/20 = 12.5 years
Q10. D, K and A can do a work in 5, 15 and 35 days respectively. They get an amount of Rs. 1054 for finishing the work working together. What is the share of K in that amount?
(a) Rs. 242
(b) Rs. 324
(c) Rs. 238
(d) Rs. 245
Sol.
D share: K share : A share
= K time × A time : D time × A time : D time × K time
= 15 × 35 : 5 × 35 : 5 × 15 = 21 : 7 : 3
So, K’s share = 1054/31 × 7 = Rs. 238
