Simple Interest
1. A sum of Rs. 2665 is lent into two parts so that the interest on the first part for 8 years at 3% per annum may be equal to the interest on the second part for 3 years at 5% per annum. Find the second sum?
A. Rs.1678
B. Rs.1640
C. Rs.2341
D. None
Answer: Option B
Explanation:
[latex]\frac{(x*8*3)}{100}[/latex] = [latex]\frac{((2665 - x)*3*5)}{100}[/latex]
[latex]\frac {24x}{100}[/latex] = [latex]\frac {39975}{100}[/latex] - [latex]\frac {15x}{100}[/latex]
39x = 39975 => x = 1025
Second sum = 2665 – 1025 = 1640
2. If A lends Rs.3500 to B at 10% per annum and B lends the same sum to C at 11.5% per annum then the gain of B in a period of 3 years is?
A. Rs.107.50
B. Rs.115.50
C. Rs.157.50
D. Rs.177.50
Answer: Option C
Explanation:
[latex]\frac{(3500*1.5*3)}{100}[/latex] = 157.50
3. If Rs.450 amount to Rs.540 in 4 years, what will it amount to in 6 years at the same rate % per annum?
A. Rs.575
B. Rs.675
C. Rs.585
D. Rs.685
Answer: Option C
Explanation:
90 = [latex]\frac{(450*4*R)}{100}[/latex]
R = 5%
I = [latex]\frac{(450*6*5)}{100}[/latex] = 135
450 + 135 = 585
Numbers
1. Five consecutive numbers add up to 335. What will be the sum of largest and smallest number?
A. 134
B. 150
C. 174
D. 226
Answer: Option A
Explanation:
Let 5 consecutive numbers be N, N+1, N+2, N+3, N+4
Sum = N+(N+1)+(N+2)+(N+3)+(N+4) = 5N + 10 = 335
∴ N (Smallest Number) = 65; Largest Number = 65 + 4 = 69
∴ 65 + 69 = 134 = Sum of largest and smallest number.
2. We reverse a number and form a new one. The old number is 45 less than new number. The sum of the digits of the old number is 9. What is the new number?
A. 36
B. 54
C. 72
D. 81
E. None of the above
Answer: Option C
Explanation:
Let the two digits be X and Y.
Let the older number be A and newer one be B.
A = 10X + Y
∴ B = 10 Y + X
From given, B = 45 + A = 45 + 10X + Y
10Y + X = 45 + 10X + Y
So, 9Y - 9X = 45;
Y - X = 5 ----------------- (1)
X + Y = 9 ---------------- (2)
Solving (1) and (2), Y = 7; X = 2
So, A = 27; B = 72
3. A farm rears geese and dogs. The headcount in the farm is 84 and the leg count is 282. How many geese are there?
Answer: Option A
Explanation:
Let geese be denoted by 'G' and Dogs by 'D'
Geese have 2 legs; Dogs have 4 legs.
Total Heads = G + D = 84 ------------------------- (1)
Total Legs = 2G + 4D = 282 --------------------- (2)
Divide equation 2 by 2, we get,
G + 2D = 141 -------------------------------------- (3)
Equation 3 - Equation 2
G + 2D - G - D = 141 - 84
∴ D = 57
So, Geese = 84 - 57 = 27
Ratio and Proportion
1. Find the 3rd proportional to 9 and 72.
A. 8
B. 216
C. 576
D. 648
Answer: Option C
Explanation:
In a:b:c, 3rd proportional is c.
a:b:c can be written as a:b::b:c
a:b::b:c can be written as[latex]\frac {a}{b}[/latex] = [latex]\frac {b}{c}[/latex]=> b[latex]^ {2}[/latex] = ac
Here, a:b:c = 9:72:c
∴ 72 x 72 = 9 x c
∴ c = [latex]\frac {72 \times 72}{9}[/latex]= 576
2. Find the 4th proportional in 7, 23 and 217?
A. 66
B. 506
C. 713
D. 961
Answer: Option C
Explanation:
In a:b::c:d , 4th proportional is d.
a:b::c:d can be written as[latex]\frac {a}{b}[/latex] =[latex]\frac {c}{d}[/latex] => d = [latex]\frac {c \times b}{a}[/latex]
Here, a:b::c:d = 7:23::217:d
∴ d = [latex]\frac {23 \times 217}{7}[/latex]= 713
3. Ram saves Rs 3395/- from his salary. He needs to pay this money as milk bill, electricity bill and mobile phone bill in the ratio 42: 32: 23. Find the money to be paid for each bill.
A. Rs 1245/-, Rs 1150/- and Rs 1000/-
B. Rs 1470/-, Rs 1120/- and Rs 805/-
C. Rs 1550/-, Rs 1235/- and Rs 610/ -
D. Rs 1764/-, Rs 1022/- and Rs 529/-
Answer: Option B
Explanation:
Common factor helps in finding actual values easily
So, take 'A' as common factor.
∴ 3 numbers will now be 42A, 32A and 23A
∴ 42A + 32A + 23 A = 3395
∴ 97A = 3395
∴ A = 35
3 parts of 3395 are
42A = 42 x 35 = 1470;
32A = 32 x 35 = 1120
23A = 23 x 35 = 805
These are the amounts to be paid.
Average
1. Kritika is on a 4 days trip with schoolmates. She had decided to keep her average expense for 4 days at Rs 80. However, at the end of the third day, she realized that her expenses on the first three days were Rs 80, Rs 90, Rs 110 and respectively. How much should she spend on 4th day to ensure that she meets her targeted average expense?
A. Rs. 40
B. Rs. 60
C. Rs. 90
D. Rs. 220
Answer: Option A
Average = [latex]\frac {Sum of Observations}{Number of Observations}[/latex]
Average = [latex]\frac {80 + 90 + 110 + 4th day}{4}[/latex] = 80
∴ 4th day expenditure = Rs. (320 - 280) = Rs 40
2. Find the average of natural numbers from 1 to 67?
A. 33.5
B. 34
C. 50
D. 67
Answer: Option B
Explanation:
Average of first n natural numbers = [latex]\frac {(n+1)}{2}[/latex]
Average of natural numbers from 1 to 65 = [latex]\frac {67+1}{2}[/latex]= 34
3. 5. The average of first 19 multiples of 12 is
A. 110
B. 110.5
C. 120
D. 220
Answer: Option C
Explanation:
Average of 'n' multiples of any number = Number x [latex]\frac {(n+1)}{2}[/latex]
IMPORTANT NOTE -
It is similar to the average of first n natural numbers.
Just multiply by the number as shown above.
Here n = 19
Average of 19 multiples of 12 = 12 x [latex]\frac {(19+1)}{2}[/latex]= 120
Problems on Ages
1. Raj's age is 4 times that of Dhiraj, his cousin. 3 years back, Raj was 5 times as old as Dhiraj. What is his present age?
A. 12 years
B. 16 years
C. 24 years
D. 48 years
Answer: Option D
Explanation:
Present day,
Let us consider Dhiraj's present age as K years
Raj's present age is 4 times Dhiraj's = 4K years
3 years ago,
Dhiraj's age = K-3 years
Raj's age = 4K-3 years.
That time, Raj's age = 5 times Dhiraj's age
∴ 4K-3 = 5(K-3)
4K-3 = 5K-15
∴ K = 12 years
Present age of Raj = 4K = 4 x 12 = 48 years.
2. Referring to his two sons Ram and Shyam, their father says - 12 years ago, Ram was 3 times as old as Shyam. 12 years later, the ratio of ages of Shyam and Ram will be 2:3. What is the present age of Ram?
A. 24 years
B. 36 years
C. 54 years
D. 144 years
Answer: Option B
Explanation:
Let us consider Shyam's present age as K years
12 years ago,
Shyam's age = K-12 years
So Ram's age is 3 times Shyam's = 3(K-12) = (3K-36) years → This age of Ram was 12 years ago
Hence, Ram's present age = (3K- 36) + 12 years = (3K-24) years
12 years later from present day,
Shyam's age = K+12 years
Ram's age = (3K-24) + 12 = (3K-12) years
Also,[latex]\frac{Shyam's age}{Ram's age}[/latex] =[latex]\frac{K+12}{3K-12}[/latex]= [latex]\frac{2}{3}[/latex]
∴ K = 20 years = Shyam's present age
∴ Ram's present age = 3K-24 = 3 x 20 - 24 = 36 years
3. The ages of 10 students and their teacher are such that the average of first 7 student's age is 15 years; average of last three student's age is 11 years. The average age of all the students and their teacher together is 15 years. Find the age of the teacher.
A. 24 years
B. 27 years
C. 30 years
D. 33 years
Answer: Option B
Explanation:
Total age of 10 students and their teacher = 11 x 15 = 165 years.
Total age of first 7 students = 7 x 15 = 105 years
Total age of 3 students = 3 x 11 = 33 years
∴ Total age of 10 students = 105 + 33 = 138 years
Age of their teacher = 165 – 138 = 27 years.
Time and Distance
1. The ratio between the speeds of the two trains is 7:8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
A. 70 km/hr
B. 75 km/hr
C. 84 km/hr
D. 87.5 km/hr
Answer: Option D
Explanation:
Let the speed of two trains be 7x and 8x km/hr.
Then, 8x = [latex]\frac {400}{4}[/latex] = 100
x = [latex]\frac {100}{8}[/latex] = 12.5
Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.
2. In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:
A. 5 kmph
B. 6 kmph
C. 6.25 kmph
D. 7.5 kmph
Answer: Option A
Explanation:
Let Abhay's speed be x km/hr.
Then, [latex]\frac {30}{x}[/latex] - [latex]\frac {30}{2x}[/latex] = 3
6x = 30
x = 5 km/hr.
3. A farmer traveled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance traveled on foot is:
A. 14 km
B. 15 km
C. 16 km
D. 17 km
Answer: Option C
Explanation:
Let the distance traveled on foot be x km.
Then, distance travelled on bicycle = (61 -x) km.
So, [latex]\frac {x}{4}[/latex] + [latex]\frac {(61 -x)}{9}[/latex] = 9
9x + 4(61 -x) = 9 x 36
5x = 80
x = 16 km.
Introduction