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RRB Online Tests
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10+10 Tests With Explanations
RRB GROUP-D & NTPC
(National Institution forTransforming India),Government of India
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Model Questions
1. The value of
is
a) 4 b) 0 c) √2 d) 3√6Sol:
Expression =
= √2 (√6 − √3) − √3(√6−√2) +√6(√3−√2)
= √12 − √6 −√18 + √6 + √18−√12= 0 Ans: b
2. If cos 2x = cos 60° cos 30° + sin60° sin 30°, then the value of x is a) 15° b) 40°c) 30° d) None of these
Sol: Cos 2x = cos 60° cos 30° + sin60° sin 30°
⇒ cos 2x =
⇒ cos 2x = cos 30°⇒ 2x = 30° ⇒ x =15° Ans: a
3. A shopkeeper allows a discountof 10% to his customers and stillgrains, 20%. Find the markedprice of the article which costsRs. 450?a) Rs. 600 b) Rs. 540c) Rs. 660 d) Rs. 580
Sol: Let the marked price of thearticle be Rs. x
∴ x ×
Ans: a4. In the figure ∆ACB − ∆APQ. If
BC = 8 cm, PQ = 4 cm, AP = 2.8cm, Find CA: a) 8 cm b) 6.5 cmc) 5.6 cm d) None of these
Sol:
Ans:c 5. When we draw the graphs of the
equations x + y = 6 and 2x + 3y= 16 on the same graph paper, thecoordinates of the point wherethe two lines intersect area) (−3, 6) b) (−2,0) c) (2, 4) d) (1, 3)
Sol: If x + y = 6m, then y = 6−xFor x = 0, y = 6 x = 2, y = 4
If 2x +3y = 16 then
For, x = 2, y = 4x = 7/2 y = 3
Ans: c6. The mean of 50 numbers is 30.
Later it was discovered that twoentries were wrongly entered as82 and 13 instead of 28 and 31.Find the correct meana) 36.12 b) 30.66c) 29.28 d) 38.21
Sol:Required average
Ans: c7. Krishna purchased a number of
articles at Rs. 10 for each and thesame number for Rs. 14 each. Hemixed them together and soldthem for Rs. 13 each. Then hisgain or loss percent is
a) Loss b) Gain
c) Loss d) Gain
Sol:Let 10 articles of each kind bebought ∴ Total cost = Rs. (10×10+14× 10) = Rs. 240Total selling price= 13×20 = Rs. 260∴ Gain percent
Ans: d8. Find a point on the y−axis which
is equidistant from the points A(6, 5) and B (−4, 3) a) (0, 9) b) (9, 0) c) (3, 0) d) (4, 0)
Sol: Let the required point be P (o, y) ∴ PA = PB
⇒ 36 + (y − 5)2 = 16 + (y − 3)2
⇒ 36 + y2 −10y +25 = 16 + y2
− 6y +9⇒ 4y =36 ⇒ y = 9∴ Required point is (0, 9).
Ans: a9. If 90% of A = 30% of B and B =
2x% of A, then the value of x is a) 450 b) 400c) 300 d) 150
Sol:⇒ 3A = B ⇒ 3A = A × 2x/100 ⇒ 300 = 2x ⇒ x = 150
Ans: d 10. A man stands at a point A on the
bank of a river and looks at thetop of a tree which is exactlyopposite to him on the otherbank. The angle of elevation is45°. He then walks 200 m at rightangles to the bank and away fromit to the point B. From B he looksat the top of the tree and theangle of elevation as 30°. Theheight of the tree is: a) 10(√3 + 1)mb) 100(√3 −1)mc) 88 (√3 +1)md) 100(√3+1)m
Sol: ∠ATE = 90°∴ ∠AET = 45°Hence AT = TE, where TErepresents the tree
Let AT = TE = x m
From ∆ BTE, = tan 30°
= 100 (√3 + 1) Ans: d11. A sum of money placed at
compound interest doubles itselfin 4 years. In how many yearswill it amount to four timesitself?a) 12 years b) 13 yearsc) 8 years d) 16 years
Sol:
∴both side square
∴ After 8 years the principle willbe 4 times itself. Ans: c
12. Ravi purchased an article markedat Rs. 21000 at 5% discount. Therate of sales tax on the article is10%. Money paid by Ravi topurchase the article is a) Rs. 19,845 b) Rs. 19,950c) Rs. 23,100 d) Rs. 21,945
Sol:Price of article after discount
Price of article after sales tax
Ans: d13. A kinetic Honda covers a
distance of 140 km in 2 hoursand 20 minutes while a carcovers the same distance in 1hour and 40 minutes. What is theratio of their speeds?a) 6 : 7 b) 3 : 7c) 4 : 7 d) 5 : 7
Sol: Speed =
∴ Required ratio
= 100 : 140 = 5 : 7 Ans: d14. In a mixture of 35 liters, the ratio
of milk and water is 4 : 1. Howmany liters of water must beadded to make the ratio 2 : 3?a) 70 b) 28c) 40 d) 35
Sol: In 35 liters of mixture,
Milk = × 35 = 28 liters
Water = 7 litersIt x liters of water be added, then
⇒ 2x + 14 =28 × 3 = 84⇒ 2x = 84 −14 =70
Ans: d15. A and B together can complete a
work in 12 days. A alone cancomplete it in 20 days. If B nowdoes the work only for half a daydaily, then the number of daysrequired to complete the work,by A and B together, is a) 15 b) 14c) 16 d) 18
Sol:B's 1 day's work
Hence, B alone will complete thework in 30 days On working for half a day, daily,time taken by B = 60 days
Ans: a16. A pump can fill a cistern in 40
minutes and another pump canempty the filled up cistern in 1hour. By mistake, with outclosing the second pump, thefirst pump is opened. The time inwhich the empty cistern will befilled up is a) 4 hours b) 1 hourc) 2 hours d) 3 hours
Sol: Part of the cistern filled in 1hour when both pumps areopened together
∴Required time = 2 hoursAns: c
17. The diameter of a roller is 84 cmand its length is 120 cm. It makes500 complete revolutions tomove once over to level aplayground. Find the area of the
playground in sq. cm.
a) 1594 sq. m b) 1580 sq. mc) 1584 sq. m. d) 1590 sq. m.
Sol:Area of the playground leveledin one revolution = 2πr × length
= 2× ×42 × 120 = 31680 sq. m.
∴ Area of the playground= (500 × 31680) sq.cm = 15840000 sq.cm= 1584 sq. metre Ans: c
18. The curved surface area of a rightcircular cylinder of radius 2 cm is264cm2. The height of thecylinder, in cm, is a) 14 b) 10.5c) 21 d) 42
Sol:Curved surface area of cylinder= 2πrh ⇒2πrh = 264
Ans: c19. A rides from his house at 12 noon
towards B's house at a speed of 8km/hour. After 1 hour, B ridesfrom his house towards A's at aspeed of 7 km/hour. If thedistance between the two housesis 68 km, the time when theymeet is a) 6 PM b) 3 PMc) 4 PM d) 5 PM
Sol: Let both meet after t hours. ∴ Distance covered by A in thours + distance covered by B in(t−1) hours = 68 km⇒ 8t + 7(t−1) = 68⇒ 8t +7t −7 = 68⇒ 15t = 68 + 7 = 75
= 5 hours i.e. at 5 p.m. Ans: d1
t75
5⇒ =
264 7
2 22 2h 21 cm⇒ = =×
× ×
2 2 h22
7264⇒ × × × =
22
7
22
7 π =
1 60 3 11 1
40 40 2 26
1
0
− = −= == −
3 1 4 1
60 60 15 15 days
+⇒ ⇒ ⇒ =
1 1
20 60⇒ +
1 1 5 3 2 1
12 20 60 60 30
−− == = =
35 liter70
s2
⇒ = =x
28 2
7 3=
+x
4
5
140km 140km:
140min utes 100min utes=
Dis tan ce
Time
19950 110 Rs. 21945Rs.
100
× ==
21000 95Rs.19950
100 Rs.
× = =
28R
1100
2 +
⇒ =
4R1
100 2 1
⇒ = +
TR
A P 1100
= +
200 200 3 1
3 1 ( 3 1)( 1)
3
× ++
⇒ =− −
=x
1
200 3⇒ =
+x
x
ET
BT
A 90 B 30
100 100
× ×=
2 2( )0 4 ( 3)⇒ + + −y
2 2( )0 6 ( 5)⇒ − + −y
20 100 18 %
240 3
×= =
260 240100
240
−= ×
18 %
3
28 %
3
28 %
3
18 %
3
30 0.72 29.236
3050
8 = + − = − =
(28 31 82 13)30
50
+ − −= +
16 2y
3
x−=
A8 AC
4C 5.6
.8cm
2⇒ =⇒ =
BC AC AB
PQ AP AQ= =
540 106 0
9 0
× =⇒ = Rs.x
9540
10⇒ =x
90 450 120
100 100
×=
3 3 3
4 4 2+ =
1 3 3 1
2 Co
2s
2 2
2× +⇒ ×=x
6( 3 2)
3 2
−−
3 2( 6 3) 4 3( 6 2)
6 3 (6 2)
− −= − +− −
3 2
3 2
−×−
4 3( 6 2) 6
( 6 2)( 6 2 ( )) 3 2
−− +
+ − +
3 2( 6 3)
( 6 3)( 6 3)
−= −+ −
3 2 4 3 6
3 6 6 2 3 2= − +
+ + +
3 2 4 3 6
3 6 6 2 3 2− +
+ + +
What is the ratio of their speeds?N. Vinaykumar Reddy
Director, IACE,
Hyderabad.
17Oòßæ§ýlÆ>»ê§Šl l Ô¶æ${MýSÐéÆý‡… l
þ Œæ l 14 l 2019ѧýlÅ Send your Feedback to [email protected]
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BV RamanaDirector, AKR Study circle
Tandur, Vikarabad
9441022571
14 .. .. ?6 14 2019 - ntnews.com nthyd.com
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Neelakantam Shekhar
Director
Shekhar’s Institute
Nalgonda
96426 76777
ANSWERS
1-1 2-3 3-3 4-2 5-1
6-3 7-2 8-2 9-3 10-2
11-3 12-3 13-3 14-4 15-1
16-2 17-2 18-1 19-1 20-3
21-1 22-2 23-2 24-1 25-2
26-4 27-3 28-4 29-4 30-3
31-3 32-2 33-4 34-1 35-3
36-2 37-4 38-2 39-1
RRB
NTPC
Special
Value of (2.3 + 1.7)2 - (2.3 - 1.7)2?
nthyd.comntnews.com 14 2019 - 7
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Þœªé-vøË™èÅ… (Geometric Progression)
a1, a2, a3, a4,......., an Íû¶C ÖÚÛ vøË™èÅ….-a1, a2, a3, a4,......., an õª øŒŠû¶u-êŸô¢ í£ë¯õªÍ÷±êŸ« a2 a3 a4 an = = = .....= = r a1 a2 a3 an1
(r Íû¶C ÖÚÛ ú‡–ô¢-ú£ÙÜuz Íô³ê¶ a1, a2, a3, a4,...., an Íû¶ vøË™èÅ…E Þœªé-vøË™èÅ… (GP), r E þ§÷«ìuEù£pAh ÍÙæ°ô¢ª.-Þœªé-vøË™-èÅ…ö˺ îμ³ë]æ¨ í£ë]Ù (a1) = a
·ôÙèÁ í£ë]Ù (a2) = îμ³ë]æ¨ í£ë]Ù r = ar
÷´èÁ í£ë]Ù (a3) = a2 r = ar r = ar2
û¦-õªÞÁ í£ë]Ù (a4) = a3 r = ar2 r = ar3
n÷ í£ë]Ù (an) = an1 r
= arn2 r = arn1
H Þœª-é-vøË™-èÅ…ö˺ þ§÷«ìu í£ë]Ù öË¶ë¯ n÷ í£ë]Ù (an)= arn1
÷«-CJ ú£÷ª-ú£uõª1.- 2, 4, 8, 16, .-.-.-.-.-.- Íû¶ Þœªé-vøË™-èÅ…ö˺ þ§÷«ìu
Eù£pAh ÓÙêŸ? 1
1) 4 2) 3 3) 2 4) - 2
þ§ëÅ]ì:- 2, 4, 8, 16, .-.-.-.- Þœªé-vøË™-èÅ…ö˺îμ³ë]æ¨ í£ë]Ù (a1) = 2 ; ·ôÙèÁ í£ë]Ù (a2) = 4
÷´èÁ í£ë]Ù (a3) = 8
a2 4þ§-÷«ìu Eù£pAh (r) = = = 2a1 2
áî¦ñª:- 3
1 3 9 272.- , , , , .-.-.-.-.- Íû¶ Þœªé-vøË™-èÅ…ö˺ 7÷ - 2 -2 -2 - 2
í£ë]Ù?729 729 243 81
1) - 2) 3) 4) - 2 3 -2 2
1 3 9 27þ§ëÅ]ì:- , , , , .-.-.-.-.- Þœªé-vøË™-èÅ…ö˺ 2 -2 -2 - 2
1 3a1 = , a2 =
2 2
3
a2 2 r = = - =- 3 r = 3
a1 1 2
7÷ í£ë]Ù (a7) = ar6
1 1= (3)6 (... a = a1 = )
2 2
729= áî¦ñª:- 1
2
3.- 3, 48 ÷ªëÅ]u 3 Þœªé-÷ª-ëÅ]u-÷«õª ÑÙç˶ þ§÷«ìuEù£pAh (r) = ?
1) 4 2) 3 3) 2 4) 6
þ§ëÅ]ì:- a, b õ ÷ªëÅ]u n Þœª-é-÷ª-ëÅ]u-÷«õª ÑÙç˶ 1
b n+1þ§÷«ìu Eù£pAh (r) = ()a
3, 48 õ ÷ªëÅ]u 3 Þœªé-÷ª-ëÅ]u-÷«õª ÑÙç˶ 1
48 3+1þ§÷«ìu Eù£pAh (r) = ()31 1 14 4 4
= (16) = (24) = 2 4
= 21 = 2 áî¦ñª:- 3
4.- ÖÚÛ Þœªé-vøË™-èÅ…-ö˺E 1÷, 3÷ í£ë¯õ îμ³êŸhÙ 15; 5÷,7÷ í£ë¯õ îμ³êŸhÙ 240 Íô³ê¶ Î vøË™èÅ…-ö˺E 9÷í£ë]Ù ÓÙêŸ? (r > 0)
1) 678 2) 768 3) 786 4) 867
þ§ëÅ]ì:- Þœªé-vøË™-èÅ…ö˺ a1 + a3 = 15
a + ar2 = 15 > (1)
a5 + a7 = 240
ar4 + ar6 = 240
r4(a + ar2) = 240
r4(15) = 240
240r4 = = 16 = 24 r = 2
15
r = 2 E s1zö˺ ví£A-¸¤-퇛úh a + ar2 = 15 a + a(2)2 = 15
a + 4a = 15 5a = 15
15 a =- = 3
5a = 3, r = 2 Í-ô³ê¶ a9 = ar8 = 3(2)8 = 3(256) = 768
áî¦ñª:- 2
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