-
Course Overview 0
No items in this section -
Numerical 48
TCS-NQT Instructions : You will be given 26 Question with a time limit of 40 mins | Questions - 26 | Time - 40 mins
-
Lecture2.1
-
Lecture2.2
-
Lecture2.3
-
Lecture2.4
-
Lecture2.5
-
Lecture2.6
-
Lecture2.7
-
Lecture2.8
-
Lecture2.9
-
Lecture2.10
-
Lecture2.11
-
Lecture2.12
-
Lecture2.13
-
Lecture2.14
-
Lecture2.15
-
Lecture2.16
-
Lecture2.17
-
Lecture2.18
-
Lecture2.19
-
Lecture2.20
-
Lecture2.21
-
Lecture2.22
-
Lecture2.23
-
Lecture2.24
-
Lecture2.25
-
Lecture2.26
-
Lecture2.27
-
Lecture2.28
-
Lecture2.29
-
Lecture2.30
-
Lecture2.31
-
Lecture2.32
-
Lecture2.33
-
Lecture2.34
-
Lecture2.35
-
Lecture2.36
-
Lecture2.37
-
Lecture2.38
-
Lecture2.39
-
Lecture2.40
-
Lecture2.41
-
Lecture2.42
-
Lecture2.43
-
Lecture2.44
-
Lecture2.45
-
Lecture2.46
-
Lecture2.47
-
Lecture2.48
-
-
Verbal 10
TCS-NQT Instructions : You will be given 24 Question with a time limit of 30 mins | Questions - 24 | Time - 15 mins + 30 mins
-
Lecture3.1
-
Lecture3.2
-
Lecture3.3
-
Lecture3.4
-
Lecture3.5
-
Lecture3.6
-
Lecture3.7
-
Lecture3.8
-
Lecture3.9
-
Lecture3.10
-
-
Aptitude Series for Placement [ Free Sample Videos ] 16
Note: TCS NQT Study Material Doesn't Contain videos we provide only Study Materials in this course , these videos are sample videos of our Aptitude Series
-
Lecture4.1
-
Lecture4.2
-
Lecture4.3
-
Lecture4.4
-
Lecture4.5
-
Lecture4.6
-
Lecture4.7
-
Lecture4.8
-
Lecture4.9
-
Lecture4.10
-
Lecture4.11
-
Lecture4.12
-
Lecture4.13
-
Lecture4.14
-
Lecture4.15
-
Lecture4.16
-
-
Reasoning 21
TCS-NQT Instructions : You will be given 30 Question with a time limit of 50 mins | Questions - 30 | Time - 50 mins
-
Lecture5.1
-
Lecture5.2
-
Lecture5.3
-
Lecture5.4
-
Lecture5.5
-
Lecture5.6
-
Lecture5.7
-
Lecture5.8
-
Lecture5.9
-
Lecture5.10
-
Lecture5.11
-
Lecture5.12
-
Lecture5.13
-
Lecture5.14
-
Lecture5.15
-
Lecture5.16
-
Lecture5.17
-
Lecture5.18
-
Lecture5.19
-
Lecture5.20
-
Lecture5.21
-
-
Programming Logic 6
TCS-NQT Instructions : You will be given 10 Question with a time limit of 15 mins | Questions - 10 | Time - 15 mins
-
Lecture6.1
-
Lecture6.2
-
Lecture6.3
-
Lecture6.4
-
Lecture6.5
-
Lecture6.6
-
-
Coding 16
TCS-NQT Instructions : You will be given 2 Question with a time limit of 15 mins and 30 mins | Questions - 2 | Time - 15 mins + 30 mins
-
Lecture7.1
-
Lecture7.2
-
Lecture7.3
-
Lecture7.4
-
Lecture7.5
-
Lecture7.6
-
Lecture7.7
-
Lecture7.8
-
Lecture7.9
-
Lecture7.10
-
Lecture7.11
-
Lecture7.12
-
Lecture7.13
-
Lecture7.14
-
Lecture7.15
-
Lecture7.16
-
-
Previous Year Paper and Solution 2
-
Lecture8.1
-
Lecture8.2
-
Δ ABC is right angled at B. If AB = 10 cm, BC = 5 cm then AC = ?
A) 3√2 cm | B) 3√5 cm |
C) 5√5 cm | D) 5√7 cm |
C) 5√5 cm
Explanation:
We have AC2 = AB2 + BC2 = 102 + 52
AC2 = 100 + 25
⇒ AC2 = 125
⇒ AC = √125 = 5√5
The angles of a traingle are 4x0, (2x - 9)0 and (7x - 6)0. The value of x is
A) 150 | B) 300 |
C) 450 | D) 600 |
A) 150
Explanation:
We have
4x0 + (2x - 9)0 + (7x - 6)0 = 1800
⇒ 4x0 + 2x0 - 90 + 7x0 - 60 = 1800
⇒ 13x0 = 1800 + 90 + 60 = 1950
⇒ x0 = 1950/13 = 150
∴ x = 150
In the given figure, what is the value of y, if 3x = 5y
A) 52.50 | B) 59.50 |
C) 67.50 | D) 72.50 |
C) 67.50
Explanation:
We have
x + y = 1800
⇒ 3x + 3y = 3 X 1800
⇒ 5y + 3y = 5400
⇒ 8y = 5400
⇒ y = 67.50
In Δ ABC, AB = AC, ∠A = x0, ∠B = (x - 15)0, then what is the value of x
A) 450 | B) 600 |
C) 700 | D) 1200 |
C) 700
Explanation:
We have
∠ B = ∠ C
⇒ (x - 15)0 = ∠ C
A/Q ∠ A + ∠ B + ∠ c = 1800
⇒ x + x - 150 + x - 150 = 1800
⇒ 3x = 1800 + 300 = 2100
⇒ x = 700
A chord AB of length 20 cm is at a distance of 4 cm from the center O of a circle. The radius of the circle is
A) 2√21 cm | B) 2√29 cm |
C) 3√12 cm | D) 3√31 cm |
B) 2√29 cm
Explanation:
We have
AB = 20 cm
OL = 4 cm
AL = 20/2 cm = 10cm
Now OA2 = AL2 + OL2 = 102 + 42 = 116 cm2
∴ OA = √116 cm2 = 2√29 cm
In the given figure, straight lines AB and CD intersect at O. If ∠x = 3∠y, then ∠y = ?
A) 200 | B) 400 |
C) 450 | D) 600 |
C) 450
Explanation:
We have
∠x + ∠y = 1800
⇒ 3∠y + ∠y = 1800
⇒ 4∠y = 1800
⇒ ∠y = 450
In the given figure, l1 || l2 and ∠A = 700, then ∠C = ?
A) 600 | B) 900 |
C) 1100 | D) 1250 |
C) 1100
Explanation:
l1 || l2
∴ ∠A = ∠B
∴ ∠B = 700
A/Q, ∠C + ∠B = 1800
⇒ ∠C = 1800 - ∠B = 1800 - 700 = 1100
In the given figure, AOB is a straight line, ∠AOC = (5x + 10)0, ∠BOC = (3x - 22)0, the value of x is
A) 240 | B) 270 |
C) 350 | D) 380 |
A) 240
Explanation:
We have
(5x + 10)0 + (3x - 22)0 = 1800
⇒ 8x = 1800 + 220 - 100
⇒ 8x = 1920
⇒ x = 240
If Δ PQR is an isosceles triangle with ∠R = 900 and RQ = 6 cm, then PQ = ?
A) 4√2 cm | B) 6√2 cm |
C) 7√2 cm | D) 9√2 cm |
B) 6√2 cm
Explanation:
We have
PR = RQ
⇒ PR = 6 cm
Now PQ2 = PR2 + RQ2 = 62 + 62 = 36 + 36 = 72
∴ PQ = √72 = 6√2 cm
In a cyclic squad ABCD, ∠A = 700, then ∠C = ?
A) 600 | B) 900 |
C) 1050 | D) 1100 |
D) 1100
Explanation:
We have
∠A + ∠C = 1800
⇒ ∠C = 1800 - ∠A = 1800 - 700 = 1100
Two poles of heights 5 m and 10 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops ?
A) 13cm | B) 16cm |
C) 19cm | D) 22cm |
A) 13cm
Explanation:
Let, AB abd CD be the poles.
Let AB = 10 m, CD = 5 m
BD = 12
AC = ?
We have AC2 = AE2 + EC2 = 52 + 122 = 169
∴ AC = √169 = 13 cm
The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively. The length of each side of the rhombus is
A)10cm | B) 15cm |
C) 18cm | D) 23cm |
B) 15cm
Explanation:
Let, ABCD be the rhombus
Let AC = 24 cm, BD = 18 cm
Now OA = ½ AC = ½ X 24 = 12 cm
OB = ½ BD = ½ X 18 = 9 cm
AB2 = AO2 + OB2 = 122 + 92 = 144 + 81 = 225
∴ AB = √225 = 15 cm
In the given figure, AOB is a diameter of the circle and CD || AB. If ∠CAB = 250, then ∠CAD = ?
A) 250 | B) 400 |
C) 450 | D) 600 |
B) 400
Explanation:
AB || CD, AC transversal
∴ ∠ACD = ∠CAB = 250
∠ACB = 900 (Same circular angle)
∴ ∠BCD = ∠ACB + ∠ACD = 900 + 250 = 1150
∠BAD + ∠BCD = 1800
⇒ ∠BAC + ∠CAD + ∠BCD = 1800
⇒ 250 + ∠CAD + 1150 = 1800
⇒ ∠CAD = 400
ABCD is a cyclic quadrilateral and AB is the diameter of the circle. If ∠CAB = 48 °, then what is the value (in degrees) of ∠ADC?
A) 52° | B) 77° |
C) 138° | D) 142° |
C) 138°
Explanation:
AB is a diameter
Therefore, ∠ACB = 90°
Also, given that, ∠CAB = 48°
∠ABC = 180° - (90° + 48°)
= 42°
ABCD is a cyclic quadrilateral
∠ADC = 180° - ∠ABC
= 180° - 42°
= 138°
In the given diagram O is the centre of the circle and CD is a tangent. ∠CAB and ∠ACD are supplementary to each other ∠OAC = 30°. Find the value of ∠OCB.
A) 30° | B) 20° |
C) 60° | D) 80° |
A) 30°
Explanation:
∠CAB and ∠ACD are supplementary to each other.
∠CAB + ∠ACD = 180°
In the given diagram, AB || CD
∠DCB = ∠ABC
Also given, ∠OAC = 30°
∠OAC = ∠OCA = 30°
Therefore, ∠AOC = 120°
∠ABC = 60°
Since, ∠DCB = ∠ABC
∠DCB = 60°
∠OCD = 90°
∠OCB = ∠OCD - ∠DCB = 90° - 60°
∠OCB = 30°.
In the given figure D, E and F are mid points of AB, AC and BC respectively. P, Q and R are mid points of DE, DF and EF. Find ratio of area of triangle PQR to that of parallelogram ADFE.
A) (1:8) | B) (1:3) |
C) (1:4) | D) (1:5) |
A) (1:8)
Explanation:
Triangle ADE, DEF, BDF and CEF are congruent to
each other. Hence share equal area of ∆ABC.
----> Area (∆ADE) = (1/4) * Area (∆ABC);
---->
Area (∆DEF) = (1/4) * Area (∆ABC);
----> Area (ADFE) = (1/2) * Area (∆ABC);
Similarly, triangle
PQR, DPQ, PER and QRF are congruent to each other and share equal are of ∆DEF
----> Area (∆PQR) =
(1/4) * Area (∆DEF);
----> = (1/4) * (1/4) * Area (∆ABC); = (1/16) * Area (∆ABC);
----> (Area
(∆PQR)/ Area (ADFE) ) = ( (1/16) * Area (∆ABC);/(1/2) * Area (∆ABC); )
----> = (1/8)
In the given figure, area of isosceles triangle ABE is 72 cm2 and BE = AB and
AB = 2 AD, AE II DC, then what is the area (in cm2) of the trapezium ABCD?
A) 108 | B) 124 |
C) 136 | D) 144 |
D) 144
Explanation:
Area of â–³ABE = 1/2 * x * x = 72
(AB = BE) = x = 12
AD = AB/2 = 6
BC = 12 + 6 = 18 cm
Area of trapezium ABCD = 1/2 * h * (AD + BC)
= 1/2 * 12 * 24
= 144
What is the average of angles x and y?
A) 80° | B) 90° |
C) 95° | D) 85° |
C) 95°
Explanation:
∠a = 40°(Vertically opposite angles)
∠b = 130°(Vertically opposite angles)
Since, the sum of angles in a trapezoid is 360°.
x + y + a + b = 360°
x + y + 40° + 130° = 360°
x + y = 190°
Average of angles x and y = 190°/2 = 95°.
In a triangle ABC, the lengths of the sides AB, AC and BC are 3, 5 and 6 cm respectively. If a point D on BC is drawn such that the line AD bisects the ∠A internally, then what is the length of BD?
A) 2 cm | B) 2.25 cm |
C) 2.5 cm | D) 3 cm |
B) 2.25 cm
Explanation:
BD/AB = DC/AC
Let BD = x
So, DC = 6 - x
x/3 = (6 - x)/5
5x = 18 -
3x
8x = 18
x = 9/4 = 2.25 cm
O is the incentre of ∆ABC and ∠A = 30° then ∠BOC is
A) 100° | B) 105° |
C) 110° | D) 90° |
B) 105°
Explanation:
∠B + ∠C = 180° - 30° = 150°
∠OBC + ∠OCB = 150°/2 = 75°
∠BOC = 180° - 75° =
105°
In a ∆ABC, incentre is O and ∠BOC = 110°, then the measure of ∠BAC is
A) 20° | B) 40° |
C) 55° | D) 110° |
B) 40°
Explanation:
110° = 90° + A/2
A/2 = 110 - 90 = 20
A = 2 * 20 = 40°