Word problems — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Word problems MCQs with step-by-step solutions (8 questions). Part of Pair of Linear Equations in Two Variables. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Word problems · easy · theory
The sum of two numbers is $35$ and their difference is $13$. The larger number is:
A. $24$ ✓ Correct
B. $11$
C. $22$
D. $13$
Solution: Let the numbers be $x$ and $y$ with $x > y$. Then $x + y = 35$ and $x - y = 13$. Adding gives $2x = 48$, so $x = 24$ and $y = 11$. The larger number is $24$.
Q2 — Word problems · easy · theory
A father's age is three times the age of his son. If the sum of their ages is $48$ years, the father's age is:
A. $12$ years
B. $24$ years
C. $36$ years ✓ Correct
D. $40$ years
Solution: Let the son's age be $s$ years; then the father's age is $3s$ years. From $s + 3s = 48$ we get $4s = 48$, so $s = 12$. The father's age is $3 \times 12 = 36$ years.
Q3 — Word problems · medium · theory
The cost of $5$ pencils and $7$ pens together is ₹50, while the cost of $7$ pencils and $5$ pens together is ₹46. The cost of one pen is:
A. ₹3
B. ₹8
C. ₹2
D. ₹5 ✓ Correct
Solution: Let a pencil cost ₹$x$ and a pen cost ₹$y$: $5x + 7y = 50$ and $7x + 5y = 46$. Adding gives $12(x + y) = 96$, so $x + y = 8$; subtracting the second from the first gives $2(y - x) = 4$, so $y - x = 2$. Solving, $y = 5$ and $x = 3$, so one pen costs ₹5.
Q4 — Word problems · medium · theory
The sum of the digits of a two-digit number is $9$. The number is $27$ more than the number obtained by reversing its digits. The number is:
A. $36$
B. $63$ ✓ Correct
C. $54$
D. $45$
Solution: Let the tens digit be $t$ and the units digit be $u$, so $t + u = 9$. The condition gives $(10t + u) - (10u + t) = 27$, i.e. $9(t - u) = 27$, so $t - u = 3$. Solving, $t = 6$ and $u = 3$, so the number is $63$; indeed $63 - 36 = 27$.
Q5 — Word problems · medium · theory
A hostel charges a fixed amount plus a daily food charge. A student who stays for $20$ days pays ₹1000, while one who stays for $26$ days pays ₹1180. The fixed charge is:
A. ₹30
B. ₹600
C. ₹1000
D. ₹400 ✓ Correct
Solution: Let the fixed charge be ₹$F$ and the daily charge be ₹$d$: $F + 20d = 1000$ and $F + 26d = 1180$. Subtracting, $6d = 180$, so $d = 30$. Then $F = 1000 - 20 \times 30 = 400$, so the fixed charge is ₹400.
Q6 — Word problems · hard · theory
A boat goes $30$ km upstream and $44$ km downstream in $10$ hours. It goes $40$ km upstream and $55$ km downstream in $13$ hours. The speed of the boat in still water is:
A. $8$ km/h ✓ Correct
B. $3$ km/h
C. $11$ km/h
D. $5$ km/h
Solution: Let the boat's speed in still water be $x$ km/h and the stream's speed be $y$ km/h. Putting $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$: $30u + 44v = 10$ and $40u + 55v = 13$. Multiplying by $4$ and $3$ respectively and subtracting gives $11v = 1$, so $v = \frac{1}{11}$, and then $u = \frac{1}{5}$. Thus $x - y = 5$ and $x + y = 11$, giving $x = 8$ km/h (and stream speed $y = 3$ km/h).
Q7 — Word problems · hard · theory
A fraction becomes $\frac{9}{11}$ if $2$ is added to both its numerator and denominator, and it becomes $\frac{5}{6}$ if $3$ is added to both. The fraction is:
A. $\frac{5}{6}$
B. $\frac{9}{11}$
C. $\frac{7}{9}$ ✓ Correct
D. $\frac{3}{5}$
Solution: Let the fraction be $\frac{x}{y}$. From $\frac{x+2}{y+2} = \frac{9}{11}$: $11x - 9y = -4$; from $\frac{x+3}{y+3} = \frac{5}{6}$: $6x - 5y = -3$. Multiplying the first by $5$ and the second by $9$ gives $55x - 45y = -20$ and $54x - 45y = -27$; subtracting, $x = 7$, and then $y = 9$. The fraction is $\frac{7}{9}$.
Q8 — Word problems · hard · theory
Five years from now, a father's age will be three times his son's age. Five years ago, the father's age was seven times his son's age. The father's present age is:
A. $35$ years
B. $45$ years
C. $50$ years
D. $40$ years ✓ Correct
Solution: Let the father's present age be $x$ years and the son's be $y$ years. Then $x + 5 = 3(y + 5)$ gives $x = 3y + 10$, and $x - 5 = 7(y - 5)$ gives $x = 7y - 30$. Equating: $3y + 10 = 7y - 30$, so $4y = 40$ and $y = 10$, giving $x = 40$. Check: in $5$ years, $45 = 3 \times 15$; and $5$ years ago, $35 = 7 \times 5$.