Algebra
Practice Questions on Algebra
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Q16) A man arranges to pay off a debt of Rs 2400 in 40 annual installments which form an AP. When 30 of the installments were repaid he dies leaving 3/8th of the debt unpaid. What is the value of the 4th installment. 1) Rs 21 2) Rs 24 3) Rs 28 4) Rs 27
Let the installments be a, a + d, a + 2d,... ATQ 20×(2a + 39d) = 2400 ⇒ 2a + 39d = 120 ...1 15×(2a + 29d) = 1500 ⇒ 2a + 29d = 100 ...2 On solving, equation 1, 2 a = 21, d = 2 4th installment = a + 3d = 27 Hence, option 4. Q17) At a certain fast food restaurant, 3 burgers, 7 shakes and 1 fries cost Rs 125. At the same place it would cost Rs 167.5 for 4 burgers, 10 shakes and 1 fries. How much would it cost 1 burger, 1 shake and 1 fries? 1) Rs 40 2) Rs 37 3) Rs 32 4) Can't be determined
ATQ 3B + 7S + 1F = 125 ...1 4B + 10S + 1F = 167.5 ...2 3(Eq 1) - 2(Eq 2) B + S + F = 40 Hence, option 4. Q18) A.P. of even consecutive integers in ascending order where first term 'n' is equal to the number of terms and thrice the number of terms in the sequence is atleast 15 more than the average of the series. Find the minimum value of 'n'. 1) 15 2) 14 3) 16 4) 18
ATQ 3n > Average of the series + 15 Sequence is n, n + 2, n + 4,...  = n + (n - 1)×2 = 3n - 2 = (n/2)(n + 3n - 2) = n(2n - 1)Average of the series  ATQ 3n > Average of the series + 15 3n > 2n - 1 + 15 n > 14 Therefore, the minimum value of n is 16 (n is even). Hence, option 3. Q19) Find the sum of n terms of the series 1.3 + 2.4 + 3.5 + 4.6 + ... n terms. 
Use substitution method Substitute n = 1 in the question. Sum of 1 term of the series = 1.3 = 3 Now, substitute n = 1 in the options. Option 1: 5/3 (eliminated) Option 2: 6 (eliminated) Option 3: 3 Option 4: 3 Now, answer is either option 3 or option 4. Substitute n = 2 in the question. Sum of 2 term of the series = 1.3 + 2.4 = 11 Now, substitute n = 2 in the options. Option 3: 11/2 (eliminated) Option 4: 11 Hence, option 4. Q20) Let Ti be the ith term of the sequence 1, 3, 6, 10, 15, 21,... and ti be the ith term of the sequence 1, 5, 14, 30, 55, 91,... Then find T15 - t15. 1) -1240 2) -1120 3) 1325 4) 1360
T15 - t15 = ∑15 - ∑152 = 120 - 1240 = -1120 Hence, option 2. |