CBSE Class 10 Mathematics Simple Equations MCQs Set 04

Answer: (a) \(x=\dfrac92\)

Explanation:
1. Multiply by the L.C.M. 66: \(11(2x-3)+6(3x-8)=2(4x+15)+33\).
2. \(22x-33+18x-48=8x+30+33\), so \(40x-81=8x+63\).
3. \(32x=144\) and \(x=\dfrac92\).

Teacher's Note:
1. Clear the fractions with the L.C.M. of all denominators.
2. Check at \(x=4.5\): LHS \(=1+\dfrac{5.5}{11}=1.5\) and RHS \(=1+0.5=1.5\).

Answer: (c) \(x=49\)

Explanation:
1. Multiply by 120: \(80(x-4)-72(2x-3)+15(3x+13)+840=0\).
2. \(80x-320-144x+216+45x+195+840=0\), so \(-19x+931=0\).
3. \(x=49\).

Teacher's Note:
1. The L.C.M. of 3, 5 and 8 is 120.
2. Check at \(x=49\): \(30-57+20+7=0\).

Answer: (a) 16 and 20

Explanation:
1. Let one part be \(x\); the other is \(36-x\).
2. \(5x=4(36-x)\) gives \(9x=144\), \(x=16\).
3. The parts are 16 and 20.

Teacher's Note:
1. Check: \(5\times16=80=4\times20\).
2. The smaller part is multiplied by the larger factor.

Answer: (b) 2 km/h

Explanation:
1. Let the stream speed be \(s\). Upstream speed \(=4-s\) and downstream speed \(=4+s\).
2. Same distance, three times the time: \(4+s=3(4-s)\).
3. \(4s=8\), so \(s=2\) km/h.

Teacher's Note:
1. Time is inversely proportional to speed for a fixed distance.
2. Check: speeds 2 and 6, ratio 1 : 3.

Answer: (a) Rs. 20

Explanation:
1. Let B give Rs. \(s\) first. A has \(100+s\), B has \(60-s\).
2. A then gives \(2s\): A has \(100-s\) and B has \(60+s\).
3. \(100-s=60+s\) so \(s=20\).

Teacher's Note:
1. Total money is Rs. 160, so each ends with Rs. 80.
2. Check: A \(=120-40=80\) and B \(=40+40=80\).

Answer: (b) \(x=a+b\)

Explanation:
1. \(ax-a^2=2ab-bx+b^2\), so \((a+b)x=a^2+2ab+b^2=(a+b)^2\).
2. Divide by \(a+b\): \(x=a+b\).

Teacher's Note:
1. Group the \(x\) terms on one side.
2. Check: \(a\cdot b=2ab-b\cdot a\) because \(x-a=b\) and \(x-b=a\).

Answer: (c) \(x=\dfrac{a+b}{2}\)

Explanation:
1. \((x+a)(x+b)-(x-a)(x-b)=2(a+b)x\).
2. So \(2(a+b)x=(a+b)^2\).
3. \(x=\dfrac{a+b}{2}\).

Teacher's Note:
1. The \(x^2\) and \(ab\) terms cancel.
2. Check with \(a=1,b=3\): \(x=2\) gives \(3\cdot5-1\cdot(-1)=16=(4)^2\).

Answer: (d) \(x=0\)

Explanation:
1. Let \(u=3^x>0\). Then \(u^2+3u=5-u\), so \(u^2+4u-5=0\).
2. \((u-1)(u+5)=0\), and \(u=-5\) is impossible, so \(u=1\).
3. \(3^x=1\) gives \(x=0\).

Teacher's Note:
1. An exponential is always positive, so reject negative values of \(u\).
2. Check: \(1+3=4=5-1\).

Answer: (d) 36

Explanation:
1. Let the number be \(10t+u\): \(7(10t+u)=4(10u+t)\), so \(66t=33u\) and \(u=2t\).
2. Digits differ by 3: \(u-t=3\), so \(t=3\) and \(u=6\).
3. The number is 36.

Teacher's Note:
1. Check: \(7\times36=252=4\times63\).
2. The reversed number 63 is larger.

Answer: (c) \(x=17,\ y=3\)

Explanation:
1. Multiply the first by 2 and the second by 3: \(6x-32y=6\) and \(6x+15y=147\).
2. Subtract: \(47y=141\), so \(y=3\).
3. \(3x=3+48=51\), so \(x=17\).

Teacher's Note:
1. Check in the second equation: \(34+15=49\).
2. Elimination is quicker than substitution here.

Answer: (c) 10 dozen oranges and 8 dozen apples

Explanation:
1. Let \(x\) dozen oranges and \(y\) dozen apples. \(7x+9y=142\) and \(8x+10y=160\).
2. Multiply the first by 8 and the second by 7: \(56x+72y=1136\) and \(56x+70y=1120\).
3. Subtract: \(2y=16\), \(y=8\); then \(7x=142-72=70\), \(x=10\).

Teacher's Note:
1. Check: \(70+72=142\) and \(80+80=160\).
2. The extra Rs. 1 per dozen on each fruit adds Rs. 18 here.