 # Maths Questions for DSSSB Exam : 3rd May 2018 (Solutions)

Q1. In the given figure BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50°, then ∠BOC =?
(a) 130°
(b) 100°
(c) 115°
(d) 120°
Q2. In the given figure, AB ∥ CD. If ∠EAB = 50° and ∠ECD = 60°, then ∠AEB= ?
(a) 50°
(b) 60°
(c) 70°
(d) 55°
Q3. In the given figure, ∠OAB=75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?
(a) 20°
(b) 25°
(c) 30°
(d) 35°
Q4. In a ∆ABC, it is given that ∠A :∠B : ∠C = 3 : 2 : 1 and CD⊥ AC. Then ∠ECD = ?
(a) 60°
(b) 45°
(c) 75°
(d) 30°
Q5. In the given figure, AB ∥ CD. If ∠ABO = 45° and ∠COD = 100°. Then ∠CDO = ?
(a) 25°
(b) 30°
(c) 35°
(d) 45°
Q6. The average of marks obtained by 120 candidates was 35. If the average of marks of passed candidates was 39 and that of failed candidates was 15, the number of candidates who passed the examination is:
(a) 100
(b) 110
(c) 80
(d) 70
Q7. The profit earned after selling an article for Rs. 1754 is the same as loss incurred after selling the article for Rs. 1492. What is the cost price of the article?
(a) Rs. 1623
(b) Rs. 1523
(c) Rs. 1689
(d) Rs. 1589
Q8. The average expenditure of a man for the first five months is Rs. 3600 and for next seven months it is Rs. 3900. If he saves Rs. 8700 during the year, his average income per month is:
(a) Rs. 4500
(b) Rs. 4200
(c) Rs. 4050
(d) Rs. 3750
Q9. The average of two numbers A and B is 20, that of B and C is 19 and of C and A is 21. What is the value of A?
(a) 24
(b) 22
(c) 20
(d) 18
Q10. 12 men can finish a project in 20 days. 18 women can finish the same project in 16 days and 24 children can finish it in 18 days. 8 women and 16 children worked for 9 days and then left. In how many days will 10 men complete the remaining project?
(a) 10 1/2
(b) 10
(c) 9
(d) 11 1/2
Solutions:
S1. Ans.(c)
Sol. ∠A + ∠B + ∠C = 180° ⇒ 50° + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 130° ⇒ 1/2∠B+1/2∠C=65°
In ∆OBC, ∠OBC + ∠OCB + ∠BOC = 180°
=1/2∠B+1/2∠C+∠BOC=180°
= 65° + ∠BOC = 180° ∴ ∠BOC = 115°
S2. Ans.(c)
Sol. Let ∠AEB = x°
Now AB ∥ CD and BC is the transversal
∴∠ABE = ∠BCD = 60° [Alt. Int. ∠s]
In ∆ABE, we have: 50° + 60° + x = 180° ∴ x = 70°
S3. Ans.(c)
Sol. In ∆OAB, we have: ∠OAB + ∠OBA + ∠AOB = 180°
∴ 75° + 55° + ∠AOB = 180° ∴ ∠AOB = 50°
∴ ∠COD = ∠AOB = 50° (vert. opp. ∠s)
In ∆OCD, we have: ∠COD + ∠OCD + ∠ODC = 180°
50° + 100° + x = 180° ∴ x = 30°
S4. Ans.(a)
Sol. We know that ∠A+∠B+∠C=180°
∴∠A=(180×3/6)^°=90°,∠B=(180×2/6)=60°and ∠C=(180×1/6)^°=30°
Ext. ∠ACE = (∠A+∠B)=(90°+60°)=150°
∴ 90° + ∠ECD = 150° ∴ ∠ECD = 60°
S5. Ans.(c)
Sol. AB ∥ CD and BOC is the transversal
∴ ∠OCD = ∠OBA = 45° [Alt. Inst. ∠S]
Now, in ∆OCD, we have :∠COD + ∠OCD + ∠CDO = 180°
∴ 100° + 45° + x = 180° ∴ x = 35°
S6. Ans.(a)
Sol. Let, the number of candidates who passed = x
Then,( 39 × x) + 15×(120-x)=120 × 35
∴ 24x = 4200 – 1800
Or, x=2400/24
x = 100.
S8. Ans.(a)
Sol. Total expenditure for the first five months = 5 × 3600 = Rs. 18000
Total expenditure for the next seven months = 7 × 3900 = Rs. 27300
Savings = Rs. 8700
Total income during the year
= 18000 + 27300 + 8700 = Rs. 54000
Average income per month =54000/12 = Rs. 4500.
S9. Ans.(b)
Sol. According to the question,
(A + B)/2=20
⇒ A + B = 40 …(i)
(B + C)/2=19
⇒ B + C = 38  …(ii)
(C + A)/2=21
⇒ C + A = 42 …(iii)
Adding (i), (ii) and (iii), we get
∴ 2A + 2B + 2C = 120
⇒ A + B + C = 60
∴ A = (A + B + C) – (B + C)  …(iv)
= 60 – 38 = 22
S10. Ans.(b)
Sol. 12M × 20 = 18W × 16 = 24C × 18
5M : 6W : 9C
Work done by 8 women and 16 children
(8W+16C)×9 = (8×5/6 M+16×5M/9)×9
=(40M/6 + 80M/9)×9 = 140 M
Remaining work = 12M × 20 – 140M = 240M – 140M = 100M
10 men will complete the work in 100/10 days = 10 days Join India's largest learning destination

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