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Masood’s age 18 years hence will be thrice his age four years ago. Find Masood’s present age ?
A. 15 years
B. 12 years
C. 54 years
D. 27 yearsLet Mudit’s present age be ‘m’ years.
m + 18 = 3(m – 4)
=> 2m = 30 => m = 15 years.Ten years ago, the age of Usman was one-third the age of Majeed at that time. The present age of Majeed is 12 years more than the present age of Usman. Find the present age of Usman ?
A. 16
B. 22
C. 43
D. 71Let the present ages of Aftab and Kashif be ‘a’ and ‘b’ respectively.
a – 10 = 1/3 (b – 10) — (1)
b = a + 12
Substituting b = a + 12 in first equation,
a – 10 = 1/3 (a + 2) => 3a – 30 = a + 2
=> 2a = 32 => a = 16.Sum of four consecutive even numbers is 292. What would be the smallest number ?
A. 23
B. 45
C. 67
D. 70Let the four consecutive even numbers be 2(x – 2), 2(x – 1), 2x, 2(x + 1)
Their sum = 8x – 4 = 292 => x = 37
Smallest number is: 2(x – 2) = 70.Cost of 16 pens and 8 pencils is Rs.352 and the cost of 4 pens and 4 pencils is Rs.96. Find the cost of each pen ?
A. Rs.76
B. Rs.67
C. Rs.32
D. Rs.2016p + 8q = 352 — (1)
4p + 4q = 96
8p + 8q = 192 — (2)
(1) – (2) => 8p = 160
=> p = 20A bag contains a total of 93 coins in the form of one rupee and 50 paise coins. If the total value of coins in the bag is Rs.56, find the number of 50 paise coins in the bag ?
A. 74
B. 76
C. 23
D. 98Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74P, Q and R have Rs.6000 among themselves. R has two-thirds of the total amount with P and Q. Find the amount with R ?
A. Rs.3600
B. Rs.4000
C. Rs.3000
D. Rs.2400Let the amount with R be Rs.r
r = 2/3 (total amount with P and Q)
r = 2/3(6000 – r) => 3r = 12000 – 2r
=> 5r = 12000 => r = 2400.Eight years ago, Zahid’s age was 4/3 times that of Jabbar. Eight years hence, Zahid’s age will be 6/5 times that of Jabbar. What is the present age of Zahid ?
A. 40 years
B. 48 years
C. 30 years
D. 32 yearsZ – 8 = 4/3 (S – 8) and Z + 8 = 6/5 (S + 8)
3/4(Z – 8) = S – 8 and 5/6(Z + 8) = S + 8
S = 3/4 (Z – 8) + 8 = 5/6 (Z + 8) – 8
=> 3/4 Z – 6 + 8 = 5/6 Z + 20/3 – 8
=> 10 – 20/3 = 10/12 Z – 9/12 Z
=> 10/3 = Z/12 => Z = 40.There are some rabbits and peacocks in a zoo. The total number of their heads is 60 and total number of their legs is 192. Find the number of total rabbits ?
A. 24
B. 56
C. 98
D. None of themone head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36.Three consecutive odd integers are in increasing order such that the sum of the last two integers is 13 more than the first integer. Find the three integers ?
A. 7, 9, 11
B. 13, 15, 17
C. 9, 11, 13
D. 11, 13, 15Let the three consecutive odd integers be x, x + 2 and x + 4 respectively.
x + 4 + x + 2 = x + 13 => x = 7
Hence three consecutive odd integers are 7, 9 and 11.A bag contains equal number of Rs.5, Rs.2 and Re.1 coins. If the total amount in the bag is Rs.1152, find the number of coins of each kind ?
A. 144
B. 72
C. 432
D. 288